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

Step 1: Identify anode, cathode, and calculate $E^\circ_{cell}$.
Since $E^\circ_{Fe^{2+}/Fe} = -0.44\,V > E^\circ_{Cr^{3+}/Cr} = -0.74\,V$, iron acts as cathode (reduction) and chromium as anode (oxidation). \[ E^\circ_{cell} = (-0.44) - (-0.74) = 0.30\,V \]
Step 2: Write balanced cell reaction and calculate $Q$.
Balancing electrons (LCM = 6): $2Cr + 3Fe^{2+} \rightarrow 2Cr^{3+} + 3Fe$ with $n = 6$. \[ Q = \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} = \frac{(0.1)^2}{(0.01)^3} = \frac{10^{-2}}{10^{-6}} = 10^4 \]
Step 3: Apply the Nernst equation.
\[ E_{cell} = E^\circ_{cell} - \frac{0.06}{n}\log Q = 0.30 - \frac{0.06}{6} \times 4 = 0.30 - 0.04 \] \[ \boxed{E_{cell} = 0.26\,V} \]