By how much is the oxidizing power of Cr2O2−7|Cr3+ couple decreased if the H+ concentration is decreased from 1M to 10−3M at 25∘C?
A
0.001V
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B
0.207V
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C
0.441V
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D
0.414V
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Solution
The correct option is C0.414V Half cell reaction is Cr2O2−7+14H++6e−→2Cr3++7H2O ECr2O2−7|Cr3+−E∘Cr2O2−7|Cr3+ =−0.0591nlog[Cr3+]2[Cr2O2−7][H+]14 =0.05916log[10−3]14=−0.414V {Option (D)}