JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Consider the following electrochemical cell at 298 K
\(Pt \left| HSnO _2^{-}( aq )\right| Sn ( OH )_6{ }^{2-}( aq )\left| Bi _2 O _3(s)\right| Bi ( s )\).
If the reaction quotient at a given time is \(10^6\), then the cell EMF ( \(E _{\text {cell }}\) ) is _______ \(\times 10^{-1} V\) (Nearest integer).
Given the standard half-cell reduction potential as
\(E _{ Bi _2 O _3 /{ Bi , OH ^{-}}^0}^0=-0.44 V\) and
\(E _{ Sn ( OH )_6^{2-} / HSnO _2^{-}, OH }^0=-0.90 V\)
- A 2
- B 3
- C 4
- D 5
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
\(E_{cell}^{\circ} = -0.44 - (-0.90)\) \(= +0.46 V\) Applying Nernst equation :- \(E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.06}{n} \log Q\) \(E _{\text {cell }}=0.46-\frac{0.06}{6} \log 10^6\) \(E _{\text {cell }}=4 \times 10^{-1}\) \(x = 4\)
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