Related Formula
Ecell = Ecell° - (0.0591)/(n) Q$$E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log Q$$
(Using 0.06 for approximation if needed, as per standard practice unless specified).
Core Logic
First, calculate E°cell$E^{\circ}_{cell}$:
E°cell = E°cathode - E°anode$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$
E°cell = (-0.44 V) - (-0.90 V) = +0.46 V$$E^{\circ}_{cell} = (-0.44\text{ V}) - (-0.90\text{ V}) = +0.46\text{ V}$$
Identify the number of transferred electrons (n$n$) for the overall reaction.
Anode reaction: Sn²⁺ arrow Sn⁴⁺ + 2e⁻$Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}$ (Tin goes from +2 in HSnO₂^-$HSnO_2^-$ to +4 in Sn(OH)₆²⁻$Sn(OH)_6^{2-}$).
Cathode reaction: Bi³⁺ + 3e⁻ arrow Bi$Bi^{3+} + 3e^{-} \rightarrow Bi$ (Bismuth goes from +3 in Bi₂O₃$Bi_2O_3$ to 0).
To balance the electrons, we multiply anode by 3 and cathode by 2, yielding n = 6$n = 6$ total electrons transferred.
Step 1: Apply Nernst Equation
Applying the Nernst equation (approximating 0.0591 to 0.06 for simplicity):
Ecell = E°cell - (0.06)/(n) ₁₀ Q$$E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log_{10} Q$$
Ecell = 0.46 - (0.06)/(6) ₁₀(10⁶)$$E_{cell} = 0.46 - \frac{0.06}{6} \log_{10}(10^6)$$
Ecell = 0.46 - 0.01 × 6$$E_{cell} = 0.46 - 0.01 \times 6$$
Ecell = 0.46 - 0.06 = 0.40 V$$E_{cell} = 0.46 - 0.06 = 0.40\text{ V}$$
Step 2: Format Output
The problem asks for the answer in the form x × 10⁻¹ V$x \times 10^{-1}\text{ V}$.
0.40 = 4 × 10⁻¹$$0.40 = 4 \times 10^{-1}$$
So, x = 4$x = 4$.
Pattern Recognition
Be extremely careful finding 'n'. You must balance the half-reactions. A +2 to +4 change (n=2) and a +3 to 0 change (n=3) resolves to a common multiple of 6.
Chapter Mix
Class 12 Chemistry: Electrochemistry