Question

Calculate emf of the following cell at \(298\,K\) : \[ Cr(s)\,|\,Cr^{3+}(aq,\;0.1\,M)\,||\,Fe^{2+}(aq,\;0.01\,M)\,|\,Fe(s) \] (Given : \[ E^\circ_{Cr^{3+}/Cr}=-0.74\,V, \] \[ E^\circ_{Fe^{2+}/Fe}=-0.44\,V, \] \[ \log 10 = 1 \] )

Hint:

For electrochemical cell numericals: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] and \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q \] Always balance the overall reaction first to determine the correct value of \(n\).

Answer 1

Concept: The emf of an electrochemical cell under non-standard conditions is calculated using the Nernst equation. The procedure involves:

• Identifying the anode and cathode.

• Calculating the standard cell potential \((E^\circ_{cell})\).

• Writing the balanced cell reaction.

• Determining the reaction quotient \((Q)\).

• Applying the Nernst equation.

Step 1: Identifying the anode and cathode. The given standard reduction potentials are: \[ E^\circ_{Cr^{3+}/Cr}=-0.74\,V \] \[ E^\circ_{Fe^{2+}/Fe}=-0.44\,V \] The electrode having the more positive reduction potential acts as the cathode. Since \[ -0.44\,V \gt -0.74\,V \] iron undergoes reduction and acts as the cathode. Therefore, \[ Fe^{2+}+2e^- \rightarrow Fe \] is the cathode reaction. Chromium undergoes oxidation and acts as the anode. \[ Cr \rightarrow Cr^{3+}+3e^- \]

Step 2: Calculating the standard emf of the cell. The standard emf is \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] Substituting the given values, \[ E^\circ_{cell} = (-0.44)-(-0.74) \] \[ E^\circ_{cell} = 0.30\,V \] Thus, \[ \boxed{E^\circ_{cell}=0.30\,V} \]

Step 3: Writing the balanced cell reaction. Oxidation half-reaction: \[ Cr \rightarrow Cr^{3+}+3e^- \] Reduction half-reaction: \[ Fe^{2+}+2e^- \rightarrow Fe \] To balance electrons, multiply: \[ Cr \rightarrow Cr^{3+}+3e^- \] by \(2\), and \[ Fe^{2+}+2e^- \rightarrow Fe \] by \(3\). Therefore, \[ 2Cr \rightarrow 2Cr^{3+}+6e^- \] \[ 3Fe^{2+}+6e^- \rightarrow 3Fe \] Adding the two equations, \[ 2Cr+3Fe^{2+} \rightarrow 2Cr^{3+}+3Fe \] Hence, \[ \boxed{n=6} \] electrons are transferred.

Step 4: Calculating the reaction quotient \(Q\). For the reaction \[ 2Cr+3Fe^{2+} \rightarrow 2Cr^{3+}+3Fe \] the solids are omitted from the expression. Therefore, \[ Q= \frac{[Cr^{3+}]^2} {[Fe^{2+}]^3} \] Substituting the concentrations, \[ Q= \frac{(0.1)^2} {(0.01)^3} \] \[ Q= \frac{10^{-2}} {10^{-6}} \] \[ Q=10^4 \] Thus, \[ \boxed{Q=10^4} \]

Step 5: Applying the Nernst equation. At \(298\,K\), \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \] Substituting the values, \[ E_{cell} = 0.30 - \frac{0.0591}{6} \log(10^4) \] Since \[ \log(10^4)=4 \] we get \[ E_{cell} = 0.30 - \frac{0.0591\times4}{6} \] \[ E_{cell} = 0.30 - 0.0394 \] \[ E_{cell} = 0.2606\,V \] \[ \boxed{E_{cell}\approx0.26\,V} \]

Step 6: Matching with the expected board answer. Many board solutions use \[ E=E^\circ-\frac{0.06}{n}\log Q \] which gives \[ E=0.30-\frac{0.06}{6}\times4 \] \[ E=0.30-0.04 \] \[ E=0.26\,V \] Thus the emf of the cell is \[ \boxed{0.26\,V} \]

Final Answer: \[ \boxed{E_{\text{cell}}=0.26\,V} \]