JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The standard electrode potential \(\left( M ^{3+} / M ^{2+}\right)\) for \(V\), \(Cr , Mn\) and \(Co\) are \(-0.26\,V ,-0.41\,V ,+1.57\,V\) and \(+1.97\,V\), respectively. The metal ions which can liberate \(H _2\) from a dilute acid are
- A \(V ^{2+}\) and \(Mn ^{2+}\)
- B \(Cr ^{2+}\) and \(CO ^{2+}\)
- C \(V ^{2+}\) and \(Cr ^{2+}\)
- D \(Mn ^{2+}\) and \(Co ^{2+}\)
Answer & Solution
Correct Answer
(C) \(V ^{2+}\) and \(Cr ^{2+}\)
Step-by-step Solution
Detailed explanation
Metal cation with \((-)\) value of reduction potential \(\left( M ^{+3} / M ^{+2}\right)\) or with \((+)\) value of oxidation potential \(\left( M ^{+2} / M ^{+3}\right)\) will liberate \(H _2\) Therefore they will reduce \(H ^{+}\) i.e \(V { }^{+2}\) and \(Cr ^{+2}\)
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