JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
At \(298\,K\), a 1 litre solution containing \(10\,m\,mol\) of \(Cr _2 O _7{ }^{2-}\) and \(100\,m\,mol\) of \(Cr ^{3+}\) shows a \(pH\) of \(3.0\). Given : \(Cr _2 O _7^{2-} \rightarrow Cr ^{3+} ; E ^0=1.330\,V\) and \(\frac{2.303 RT }{ F }=0.059\,V\) The potential for the half cell reaction is \(x \times 10^{-3}\,V\). The value of \(x\) is \(........\)
- A \(916\)
- B \(915\)
- C \(917\)
- D \(914\)
Answer & Solution
Correct Answer
(C) \(917\)
Step-by-step Solution
Detailed explanation
\(Cr _2 O _7^{2-}+14 H ^{+}+6 e ^{-} \rightarrow 2 Cr ^{3+}+7 H _2 O\) \(E =1.33-\frac{0.059}{6} \log \frac{(0.1)^2}{\left(10^{-2}\right)\left(10^{-3}\right)^{14}}\) \(E =1.33-\frac{0.059}{6} \times 42=0.917\) \(E =917 \times 10^{-3}\) \(x =917\)
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