{"id":6157,"date":"2026-10-26T14:10:08","date_gmt":"2026-10-26T08:40:08","guid":{"rendered":"https:\/\/www.catmock.com\/blog\/?p=6157"},"modified":"2026-10-26T14:10:09","modified_gmt":"2026-10-26T08:40:09","slug":"for-a-half-cell-reaction-half-cl2-nernst-answer","status":"publish","type":"post","link":"https:\/\/www.catmock.com\/blog\/for-a-half-cell-reaction-half-cl2-nernst-answer\/","title":{"rendered":"For a Half Cell Reaction \u00bdCl\u2082: Nernst Equation, Answer and Steps"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\"><strong>For a half cell reaction \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b, the Nernst equation at 298 K is E = 1.36 \u2212 0.0591 log ([Cl\u207b] \/ P(Cl\u2082)^\u00bd) volts.<\/strong> Here n = 1, E\u00b0 = +1.36 V and Q = [Cl\u207b] \/ \u221aP(Cl\u2082). A higher [Cl\u207b] lowers E, while a higher chlorine pressure raises it.<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\"><strong>Quick Answer:<\/strong> E = E\u00b0 \u2212 (0.0591\/n) log Q, with E\u00b0 = +1.36 V, n = 1 and Q = [Cl\u207b] \/ (P_Cl\u2082)^\u00bd. At standard conditions (1 M Cl\u207b, 1 atm Cl\u2082), E = +1.36 V.<\/p>\n<\/blockquote>\n\n\n\n<h2 class=\"wp-block-heading\">Key Takeaways<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The half-cell reaction \u00bdCl\u2082(g) + e\u207b \u2192 Cl\u207b(aq) describes reduction at the Cl\u2082\/Cl\u207b electrode.<\/li>\n\n\n\n<li>Its standard electrode potential is about +1.36 V at 298 K.<\/li>\n\n\n\n<li>The number of electrons is n = 1, because only half a mole of Cl\u2082 is written.<\/li>\n\n\n\n<li>The reaction quotient is Q = [Cl\u207b] \/ (P_Cl\u2082)^\u00bd.<\/li>\n\n\n\n<li>Writing \u00bdCl\u2082 or Cl\u2082 changes n and Q, but never the electrode potential E.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">What does the half-cell reaction \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b mean?<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>It means half a mole of chlorine gas gains one electron to form one mole of chloride ions.<\/strong> This is a reduction reaction, so the electrode acts as a cathode when connected to a weaker oxidising half-cell.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A <strong>half-cell<\/strong> is one electrode dipped in its electrolyte. In this case, an inert platinum wire sits in a chloride solution while chlorine gas bubbles over it. Platinum only carries electrons. It does not take part in the reaction.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">How is the chlorine electrode written?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The cell notation is Pt | Cl\u2082(g, P) | Cl\u207b(aq, c). The vertical bars show phase boundaries. The gas and the ions meet at the platinum surface, where electrons move in or out.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why is the reaction written with \u00bd?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Writing \u00bdCl\u2082 gives exactly one electron per chloride ion formed. Many textbooks and exam questions use this form because it keeps n = 1 and makes the Q expression short. It is the same chemistry as Cl\u2082 + 2e\u207b \u2192 2Cl\u207b, only scaled down.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">What is the standard electrode potential of the Cl\u2082\/Cl\u207b electrode?<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The standard electrode potential of the Cl\u2082\/Cl\u207b electrode is about +1.36 V at 298 K, 1 M Cl\u207b and 1 atm Cl\u2082.<\/strong> Textbook tables list 1.36 V, and some precise tables give 1.358 V.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A <strong>standard electrode potential (E\u00b0)<\/strong> is the reduction potential measured against the standard hydrogen electrode (SHE) under standard conditions. The SHE is assigned 0.00 V.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A large positive E\u00b0 means chlorine is a strong oxidising agent. It grabs electrons easily. That is why chlorine can oxidise bromide and iodide ions.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Does E\u00b0 change when the equation is multiplied?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">No. E\u00b0 is an intensive property. It does not depend on how much substance is written in the equation. So E\u00b0 stays +1.36 V for both \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b and Cl\u2082 + 2e\u207b \u2192 2Cl\u207b.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Assumptions used in this article:<\/strong> temperature is 298 K, chlorine behaves as an ideal gas, and concentrations replace activities. The values given are standard-condition values.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">What is the Nernst equation for \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b?<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The Nernst equation for the half-cell \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b at 298 K is E = 1.36 \u2212 0.0591 log ([Cl\u207b] \/ P_Cl\u2082^\u00bd).<\/strong> It links the actual electrode potential to the concentration of chloride ions and the pressure of chlorine gas.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The <strong>Nernst equation<\/strong> tells you how the electrode potential shifts when conditions move away from standard.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What is the general form?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The general form is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>E = E\u00b0 \u2212 (RT \/ nF) ln Q<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Here, R is the gas constant, T is temperature in kelvin, n is the number of electrons, F is the Faraday constant and Q is the reaction quotient.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What is the form at 298 K?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">At 298 K, the factor 2.303RT\/F equals about 0.0591 V. Converting ln to log gives:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>E = E\u00b0 \u2212 (0.0591 \/ n) log Q<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What is Q for this half cell?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Q is the products over the reactants, each raised to its coefficient. Pure solids and liquids are left out. For \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Q = [Cl\u207b] \/ (P_Cl\u2082)^\u00bd<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substituting n = 1 and E\u00b0 = 1.36 V gives the final working equation:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>E = 1.36 \u2212 0.0591 log ([Cl\u207b] \/ \u221aP_Cl\u2082)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This can also be written as E = 1.36 + 0.02955 log P_Cl\u2082 \u2212 0.0591 log [Cl\u207b]. Both forms give the same answer.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Nernst equation variables<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><th>Symbol<\/th><th>Meaning<\/th><th>Value or unit<\/th><\/tr><tr><td>E<\/td><td>Electrode potential under the given conditions<\/td><td>Volts (V)<\/td><\/tr><tr><td>E\u00b0<\/td><td>Standard electrode potential of Cl\u2082\/Cl\u207b<\/td><td>+1.36 V<\/td><\/tr><tr><td>R<\/td><td>Universal gas constant<\/td><td>8.314 J K\u207b\u00b9 mol\u207b\u00b9<\/td><\/tr><tr><td>T<\/td><td>Absolute temperature<\/td><td>298 K (25 \u00b0C)<\/td><\/tr><tr><td>F<\/td><td>Faraday constant<\/td><td>96,485 C mol\u207b\u00b9<\/td><\/tr><tr><td>n<\/td><td>Electrons transferred in the written equation<\/td><td>1 for the \u00bdCl\u2082 form<\/td><\/tr><tr><td>Q<\/td><td>Reaction quotient<\/td><td>[Cl\u207b] \/ (P_Cl\u2082)^\u00bd<\/td><\/tr><tr><td>2.303RT\/F<\/td><td>Combined factor at 298 K<\/td><td>About 0.0591 V<\/td><\/tr><tr><td>P_Cl\u2082<\/td><td>Partial pressure of chlorine<\/td><td>Atmospheres (atm)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Note: IUPAC now defines the standard pressure as 1 bar. The difference from 1 atm is tiny, so exam problems treat them as equal.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Why does the coefficient \u00bd matter?<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The \u00bd matters because it sets n = 1 and puts a square root on the chlorine pressure in Q.<\/strong> It does not change E\u00b0 or the final potential E.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Students often lose marks here. They write Q with P_Cl\u2082 to the power 1 but use n = 1, or the reverse. Both errors give wrong answers.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">The rule for coefficients<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each coefficient in the balanced half-reaction becomes an exponent in Q.<\/li>\n\n\n\n<li>The number of electrons n must match the same written equation.<\/li>\n\n\n\n<li>E and E\u00b0 are unchanged when you multiply the equation.<\/li>\n\n\n\n<li>\u0394G and the log Q term scale together with n, so the final E is the same.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Does \u0394G change?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Yes. Since \u0394G = \u2212nFE, doubling the equation doubles n and doubles \u0394G. But E stays the same. This is the clearest way to see why E is independent of the way you write the equation.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Comparing the two ways of writing the chlorine half-cell<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><th>Feature<\/th><th>\u00bdCl\u2082 + e\u207b \u2192 Cl\u207b<\/th><th>Cl\u2082 + 2e\u207b \u2192 2Cl\u207b<\/th><\/tr><tr><td>n (electrons)<\/td><td>1<\/td><td>2<\/td><\/tr><tr><td>Q expression<\/td><td>[Cl\u207b] \/ (P_Cl\u2082)^\u00bd<\/td><td>[Cl\u207b]\u00b2 \/ P_Cl\u2082<\/td><\/tr><tr><td>E\u00b0<\/td><td>+1.36 V<\/td><td>+1.36 V<\/td><\/tr><tr><td>Nernst form at 298 K<\/td><td>E = 1.36 \u2212 0.0591 log Q<\/td><td>E = 1.36 \u2212 (0.0591\/2) log Q<\/td><\/tr><tr><td>\u0394G\u00b0<\/td><td>\u2212F(1.36)<\/td><td>\u22122F(1.36)<\/td><\/tr><tr><td>Final E for the same conditions<\/td><td>Same<\/td><td>Same<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><em>What changes: n, Q and \u0394G. What does not change: E\u00b0 and E.<\/em><\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Step-by-step answer to the typical question<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A commonly asked version of this question asks for the electrode potential of the Cl\u2082\/Cl\u207b half cell at a given [Cl\u207b] and chlorine pressure.<\/strong> The exact wording differs between books and exams, so the values below are chosen for illustration.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Problem:<\/strong> Calculate the electrode potential of the half cell Pt | Cl\u2082(10 atm) | Cl\u207b(0.1 M) at 298 K for the reaction \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b. Take E\u00b0 = +1.36 V.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Write the half-reaction.<\/strong> \u00bdCl\u2082(g) + e\u207b \u2192 Cl\u207b(aq). So n = 1.<\/li>\n\n\n\n<li><strong>Write Q.<\/strong> Q = [Cl\u207b] \/ (P_Cl\u2082)^\u00bd.<\/li>\n\n\n\n<li><strong>Substitute the values.<\/strong> Q = 0.1 \/ (10)^\u00bd = 0.1 \/ 3.162 = 0.03162.<\/li>\n\n\n\n<li><strong>Find log Q.<\/strong> log(0.03162) = \u22121.5.<\/li>\n\n\n\n<li><strong>Apply the Nernst equation.<\/strong> E = E\u00b0 \u2212 (0.0591 \/ 1) log Q = 1.36 \u2212 0.0591 \u00d7 (\u22121.5).<\/li>\n\n\n\n<li><strong>Calculate.<\/strong> E = 1.36 + 0.0887 = 1.4487 V.<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer: E \u2248 1.449 V.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Quick check using the other form<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Write the reaction as Cl\u2082 + 2e\u207b \u2192 2Cl\u207b. Then n = 2 and Q = (0.1)\u00b2 \/ 10 = 0.001. So E = 1.36 \u2212 (0.0591 \/ 2) \u00d7 log(0.001) = 1.36 + 0.0887 = 1.4487 V. The same answer. This proves that the \u00bd form and the full form agree.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">A second solved example<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Problem:<\/strong> Find E for the Cl\u2082\/Cl\u207b half cell when [Cl\u207b] = 0.010 M and P_Cl\u2082 = 4 atm at 298 K.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Reaction and n.<\/strong> \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b, n = 1.<\/li>\n\n\n\n<li><strong>Q.<\/strong> Q = 0.010 \/ (4)^\u00bd = 0.010 \/ 2 = 0.005.<\/li>\n\n\n\n<li><strong>log Q.<\/strong> log(0.005) = \u22122.301.<\/li>\n\n\n\n<li><strong>E.<\/strong> E = 1.36 \u2212 0.0591 \u00d7 (\u22122.301) = 1.36 + 0.1360.<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer: E \u2248 1.496 V.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What does the result tell us?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The potential is higher than 1.36 V. Both changes push it up: a lower [Cl\u207b] and a higher chlorine pressure. Low [Cl\u207b] means less product, so the reduction is more favourable. High pressure means more reactant.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What if the cell is at standard conditions?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">If [Cl\u207b] = 1 M and P_Cl\u2082 = 1 atm, then Q = 1 and log Q = 0. So E = E\u00b0 = 1.36 V. Use this as a quick sanity check in any problem.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">How do [Cl\u207b] and P(Cl\u2082) change the potential?<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A higher [Cl\u207b] lowers the electrode potential, and a higher P(Cl\u2082) raises it.<\/strong> This follows from Le Chatelier&#8217;s principle and from the sign in the Nernst equation.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Chloride is a product, so more of it makes the forward reduction less favourable. Chlorine gas is a reactant, so more of it makes reduction more favourable.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Table 2: Effect of changing concentration or pressure on E<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><th>Change (other factors constant)<\/th><th>Effect on Q<\/th><th>Effect on log Q<\/th><th>Effect on E<\/th><\/tr><tr><td>[Cl\u207b] increases<\/td><td>Q increases<\/td><td>Increases<\/td><td>E decreases<\/td><\/tr><tr><td>[Cl\u207b] decreases<\/td><td>Q decreases<\/td><td>Decreases<\/td><td>E increases<\/td><\/tr><tr><td>P(Cl\u2082) increases<\/td><td>Q decreases<\/td><td>Decreases<\/td><td>E increases<\/td><\/tr><tr><td>P(Cl\u2082) decreases<\/td><td>Q increases<\/td><td>Increases<\/td><td>E decreases<\/td><\/tr><tr><td>[Cl\u207b] \u00d710<\/td><td>Q \u00d710<\/td><td>+1<\/td><td>E falls by 0.0591 V<\/td><\/tr><tr><td>P(Cl\u2082) \u00d7100<\/td><td>Q \u00f710<\/td><td>\u22121<\/td><td>E rises by 0.0591 V<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\">Which factor has a bigger effect?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Chloride concentration has a stronger effect per tenfold change. A tenfold change in [Cl\u207b] shifts E by 0.0591 V. A tenfold change in pressure shifts E by only 0.02955 V, because of the square root.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Common mistakes students make<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The most common mistake is using the wrong n or the wrong exponent for chlorine pressure.<\/strong> Here is how students usually get this wrong, and how to avoid it.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Using n = 2 with the \u00bd equation.<\/strong> If the equation has \u00bdCl\u2082, n is 1.<\/li>\n\n\n\n<li><strong>Forgetting the square root.<\/strong> With \u00bdCl\u2082, the pressure term is P^\u00bd, not P.<\/li>\n\n\n\n<li><strong>Inverting Q.<\/strong> For reduction, Q = products \/ reactants, so [Cl\u207b] goes on top.<\/li>\n\n\n\n<li><strong>Using 0.059 for everything.<\/strong> The factor 0.0591 is only valid at 298 K.<\/li>\n\n\n\n<li><strong>Mixing up the sign.<\/strong> The Nernst equation has a minus sign before the log term when Q is products over reactants.<\/li>\n\n\n\n<li><strong>Putting the solid or platinum in Q.<\/strong> Pure solids and the inert electrode are left out.<\/li>\n\n\n\n<li><strong>Using Celsius.<\/strong> Always convert temperature to kelvin when using the general RT\/nF form.<\/li>\n<\/ul>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\"><strong>Practical tip:<\/strong> Before you finish any Nernst problem, ask one question. Should E move up or down compared with E\u00b0? If your answer goes the wrong way, recheck Q.<\/p>\n<\/blockquote>\n\n\n\n<h2 class=\"wp-block-heading\">How this concept appears in competitive exams<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Exams test this concept through direct calculations, concentration effects and links between E, \u0394G and K.<\/strong> It is a regular topic in the electrochemistry chapter of Class 12, JEE, NEET and CUET preparation.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Typical question patterns include:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Finding E of a chlorine electrode at a given [Cl\u207b] and pressure.<\/li>\n\n\n\n<li>Finding the emf of a full cell that pairs the Cl\u2082\/Cl\u207b electrode with another half cell.<\/li>\n\n\n\n<li>Using E\u00b0 values to find the E\u00b0 of a related couple, such as OCl\u207b\/Cl\u207b and Cl\u207b\/\u00bdCl\u2082, by adding half-reactions.<\/li>\n\n\n\n<li>Finding K or K_sp from E\u00b0 using E\u00b0 = (0.0591\/n) log K.<\/li>\n\n\n\n<li>Judging how E changes when pressure or concentration is varied.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">In problems that reverse the reaction (Cl\u207b \u2192 \u00bdCl\u2082 + e\u207b), the sign of E\u00b0 flips to \u22121.36 V. Be careful to keep the equation direction consistent with your Q.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Tips for solving faster<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Write both half-reactions as reductions first.<\/li>\n\n\n\n<li>Pick n from the equation exactly as written.<\/li>\n\n\n\n<li>Build Q straight from the coefficients.<\/li>\n\n\n\n<li>Estimate the direction of change before you calculate.<\/li>\n\n\n\n<li>Round only at the final step.<\/li>\n<\/ol>\n\n\n\n<h2 class=\"wp-block-heading\">PDF-style quick revision notes<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Use this section as a printable formula sheet.<\/strong> Save the page as PDF from your browser, or copy these notes into your own revision file.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Formula sheet<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Reaction: \u00bdCl\u2082(g) + e\u207b \u2192 Cl\u207b(aq)<\/li>\n\n\n\n<li>E\u00b0 (Cl\u2082\/Cl\u207b) = +1.36 V<\/li>\n\n\n\n<li>n = 1<\/li>\n\n\n\n<li>Q = [Cl\u207b] \/ (P_Cl\u2082)^\u00bd<\/li>\n\n\n\n<li>E = E\u00b0 \u2212 (0.0591\/n) log Q (298 K)<\/li>\n\n\n\n<li>E = 1.36 \u2212 0.0591 log ([Cl\u207b] \/ \u221aP_Cl\u2082)<\/li>\n\n\n\n<li>\u0394G = \u2212nFE and \u0394G\u00b0 = \u2212nFE\u00b0<\/li>\n\n\n\n<li>E\u00b0 = (0.0591\/n) log K (at equilibrium)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Memory aids<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Chloride up, potential down.<\/li>\n\n\n\n<li>Chlorine pressure up, potential up.<\/li>\n\n\n\n<li>Coefficient in the equation becomes the exponent in Q.<\/li>\n\n\n\n<li>n follows the equation as written. E does not.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Five-line summary<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The Cl\u2082\/Cl\u207b electrode is a gas electrode on inert platinum.<\/li>\n\n\n\n<li>Its standard potential is +1.36 V.<\/li>\n\n\n\n<li>The \u00bd form has n = 1.<\/li>\n\n\n\n<li>Q uses the square root of chlorine pressure.<\/li>\n\n\n\n<li>The final E is the same for any way of writing the equation.<\/li>\n<\/ol>\n\n\n\n<h2 class=\"wp-block-heading\">FAQs<\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">What is the half-cell reaction \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">It is the reduction of chlorine gas to chloride ions at the Cl\u2082\/Cl\u207b electrode. Half a mole of Cl\u2082 takes up one mole of electrons to give one mole of Cl\u207b. It is a reduction reaction with n = 1.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What is the standard electrode potential of the chlorine electrode?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The standard electrode potential of the Cl\u2082\/Cl\u207b electrode is about +1.36 V at 298 K. This is measured against the standard hydrogen electrode under 1 M Cl\u207b and 1 atm Cl\u2082. A large positive value shows chlorine is a strong oxidising agent.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What is the Nernst equation for \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">At 298 K it is E = 1.36 \u2212 0.0591 log ([Cl\u207b] \/ P_Cl\u2082^\u00bd). The general form is E = E\u00b0 \u2212 (RT\/nF) ln Q. Here n = 1 and Q = [Cl\u207b] \/ (P_Cl\u2082)^\u00bd.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why is n equal to 1 in this half reaction?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The equation shows one electron on the reactant side. Since only half a mole of Cl\u2082 is written, one chloride ion forms per electron gained. So n = 1. In the Cl\u2082 + 2e\u207b \u2192 2Cl\u207b form, n would be 2.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Does the \u00bd coefficient change the electrode potential?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">No. The electrode potential E is an intensive property, so it does not depend on how the equation is scaled. The \u00bd changes n, Q and \u0394G, but E and E\u00b0 stay the same.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What happens to E if the chloride concentration increases?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">E decreases. Chloride is a product, so a higher [Cl\u207b] increases Q and makes log Q larger. Because the Nernst equation subtracts this term, the electrode potential falls. A tenfold rise lowers E by 0.0591 V at 298 K.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What happens to E if chlorine pressure increases?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">E increases. Chlorine gas is a reactant, so higher pressure lowers Q, which makes log Q smaller. The Nernst equation then gives a higher potential. A hundredfold rise in pressure raises E by 0.0591 V in the \u00bdCl\u2082 form.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why is the value 0.0591 V used?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">It comes from 2.303RT\/F at 298 K. Using R = 8.314 J K\u207b\u00b9 mol\u207b\u00b9, T = 298 K and F = 96,485 C mol\u207b\u00b9 gives about 0.0591 V. Some books round it to 0.059 V. Use the value your exam specifies.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Is there a downloadable PDF for this topic?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This page includes a printable formula sheet and revision notes. You can save the page as PDF from your browser&#8217;s print menu and keep it for quick revision before tests.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What is the difference between E\u00b0 and E?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">E\u00b0 is the potential under standard conditions (1 M, 1 atm, 298 K). E is the potential under any actual conditions. The Nernst equation connects the two, and they are equal only when Q = 1.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Conclusion<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">For a half cell reaction \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b, remember three facts: E\u00b0 = +1.36 V, n = 1 and Q = [Cl\u207b] \/ \u221aP_Cl\u2082. Put these into E = E\u00b0 \u2212 (0.0591\/n) log Q, and you can solve most problems on the chlorine electrode.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The best next step is practice. Try changing the values in the two solved examples and check that the \u00bd form and the full form agree every time.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Reliable sources for further reading<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/en.wikipedia.org\/wiki\/Nernst_equation\" target=\"_blank\" rel=\"noopener\">Nernst equation overview (Wikipedia)<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/archive.nptel.ac.in\/content\/storage2\/courses\/104106129\/module1\/lec2.pdf\" target=\"_blank\" rel=\"noopener\">NPTEL: Electrode potentials and applications (PDF)<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/ncert.nic.in\/textbook.php\" target=\"_blank\" rel=\"noopener\">NCERT textbooks<\/a> (Class 12 Chemistry, Electrochemistry chapter)<\/li>\n\n\n\n<li><a href=\"https:\/\/goldbook.iupac.org\/\" target=\"_blank\" rel=\"noopener\">IUPAC Gold Book<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/chem.libretexts.org\/\" target=\"_blank\" rel=\"noopener\">LibreTexts Chemistry<\/a><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>You may also like<\/strong> : <strong><a href=\"https:\/\/www.catmock.com\/\">Cat Mock<\/a><\/strong>, <strong><a href=\"https:\/\/www.catmock.com\/\">Catmock<\/a><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>For a half cell reaction \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b, the Nernst equation at 298 K is E = 1.36 \u2212 0.0591 log ([Cl\u207b] \/ P(Cl\u2082)^\u00bd) volts. Here n = 1, E\u00b0 = +1.36 V and Q = [Cl\u207b] \/ \u221aP(Cl\u2082). A higher [Cl\u207b] lowers E, while a higher chlorine pressure raises it. Quick Answer: [&hellip;]<\/p>\n","protected":false},"author":6,"featured_media":6158,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ocean_front_end_style_editor":"no","ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"","ocean_second_sidebar":"","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"","ocean_custom_header_template":"","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"","ocean_menu_typo_font_family":"","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"default","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"default","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":"","ocean_post_oembed":"","ocean_post_self_hosted_media":"","ocean_post_video_embed":"","ocean_link_format":"","ocean_link_format_target":"self","ocean_quote_format":"","ocean_quote_format_link":"post","ocean_gallery_link_images":"on","ocean_gallery_id":[],"footnotes":""},"categories":[835],"tags":[],"class_list":["post-6157","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-for-a-half-cell-reaction-cl","entry","has-media"],"featured_image_src":"https:\/\/www.catmock.com\/blog\/wp-content\/uploads\/2026\/10\/For-a-Half-Cell-Reaction-\u00bdCl\u2082.jpg","author_info":{"display_name":"Yatresh Sharma","author_link":"https:\/\/www.catmock.com\/blog\/author\/yatresh-sharma\/"},"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.7 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>For a Half Cell Reaction \u00bdCl\u2082: Nernst Answer &amp; Steps<\/title>\n<meta name=\"description\" content=\"Solve for a half cell reaction \u00bdCl\u2082 + e\u207b \u2192 Cl\u207b with the Nernst equation. 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