JEE Challenger
More from Electrochemistry
ChemMedMain 2026 • 4th Apr - Shift 2

Calculation of Metal Ion Charge in Electrochemical Cell Using Nernst Equation

An electrochemical cell, consist of the following two redox couples, Mx+(aq)/M(s)\text{M}^{x+}(\text{aq})/\text{M}(\text{s}) [Ered⊖=+0.15 VE_{\text{red}}^{\ominus} = +0.15\text{ V}] and Fe3+(aq)/Fe(s)\text{Fe}^{3+}(\text{aq})/\text{Fe}(\text{s}) [Ered⊖=−0.036 VE_{\text{red}}^{\ominus} = -0.036\text{ V}]. The cell EMF (EcellE_{\text{cell}}) is recorded to be 0.2057 V0.2057\text{ V}. If the reaction quotient of the electrochemical reaction is found to be 10−210^{-2}, then the value of xx is ______.(Nearest integer)

[Given : M is a p-block metal and 2.303RTF=0.059 V\frac{2.303\text{RT}}{\text{F}} = 0.059\text{ V}]

Official Numerical Answer2

Topics & Concepts

ElectrochemistryElectrochemical CellNernst Equation

Step-by-Step Solution

To find the value of xx, we apply the Nernst equation to the given electrochemical cell.

Step 1: Identify the Cathode and Anode

The electrode with the higher standard reduction potential acts as the cathode (where reduction occurs), and the electrode with the lower standard reduction potential acts as the anode (where oxidation occurs).

Given:

  • Ered⊖(Mx+/M)=+0.15 VE_{\text{red}}^{\ominus}(\text{M}^{x+}/\text{M}) = +0.15\text{ V}
  • Ered⊖(Fe3+/Fe)=−0.036 VE_{\text{red}}^{\ominus}(\text{Fe}^{3+}/\text{Fe}) = -0.036\text{ V}

Since Ered⊖(Mx+/M)>Ered⊖(Fe3+/Fe)E_{\text{red}}^{\ominus}(\text{M}^{x+}/\text{M}) > E_{\text{red}}^{\ominus}(\text{Fe}^{3+}/\text{Fe}),

  • Cathode (Reduction): Mx+(aq)+xe−→M(s)\text{M}^{x+}(\text{aq}) + x e^{-} \rightarrow \text{M}(\text{s})
  • Anode (Oxidation): Fe(s)→Fe3+(aq)+3e−\text{Fe}(\text{s}) \rightarrow \text{Fe}^{3+}(\text{aq}) + 3 e^{-}

Step 2: Calculate Standard Cell EMF (Ecell⊖E_{\text{cell}}^{\ominus})

Ecell⊖=Ecathode⊖−Eanode⊖E_{\text{cell}}^{\ominus} = E_{\text{cathode}}^{\ominus} - E_{\text{anode}}^{\ominus} Ecell⊖=0.15 V−(−0.036 V)=0.186 VE_{\text{cell}}^{\ominus} = 0.15\text{ V} - (-0.036\text{ V}) = 0.186\text{ V}

Step 3: Determine Total Number of Electrons (nn)

To balance the overall cell reaction, multiply the cathode reaction by 33 and the anode reaction by xx:

Cathode: 3Mx+(aq)+3xe−→3M(s)\text{Cathode: } 3\text{M}^{x+}(\text{aq}) + 3x e^{-} \rightarrow 3\text{M}(\text{s}) Anode: xFe(s)→xFe3+(aq)+3xe−\text{Anode: } x\text{Fe}(\text{s}) \rightarrow x\text{Fe}^{3+}(\text{aq}) + 3x e^{-}

Overall Cell Reaction: 3Mx+(aq)+xFe(s)→3M(s)+xFe3+(aq)3\text{M}^{x+}(\text{aq}) + x\text{Fe}(\text{s}) \rightarrow 3\text{M}(\text{s}) + x\text{Fe}^{3+}(\text{aq})

The total number of moles of electrons transferred in the balanced cell reaction is: n=3xn = 3x

Step 4: Apply Nernst Equation

The Nernst equation for the cell reaction is: Ecell=Ecell⊖−2.303RTnFlog⁡10QE_{\text{cell}} = E_{\text{cell}}^{\ominus} - \frac{2.303 RT}{n F} \log_{10} Q

Given values:

  • Ecell=0.2057 VE_{\text{cell}} = 0.2057\text{ V}
  • Ecell⊖=0.186 VE_{\text{cell}}^{\ominus} = 0.186\text{ V}
  • 2.303RTF=0.059 V\frac{2.303 RT}{F} = 0.059\text{ V}
  • Q=10−2Q = 10^{-2}
  • n=3xn = 3x

Substitute these values into the equation: 0.2057=0.186−0.0593xlog⁡10(10−2)0.2057 = 0.186 - \frac{0.059}{3x} \log_{10}(10^{-2})

Since log⁡10(10−2)=−2\log_{10}(10^{-2}) = -2: 0.2057−0.186=−0.0593x(−2)0.2057 - 0.186 = -\frac{0.059}{3x} (-2) 0.0197=0.1183x0.0197 = \frac{0.118}{3x}

Solving for 3x3x: 3x=0.1180.0197≈5.9898≈63x = \frac{0.118}{0.0197} \approx 5.9898 \approx 6

x=63=2x = \frac{6}{3} = 2

Thus, the value of xx is 2.

Calculation of Metal Ion Charge in Electrochemical Cell Using Nernst Equation | Chemistry PYQ Solution - JEE Challenger