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Given: (i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V` (ii) `Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V` Electrode potential, `E^(@)` for the reaction, `Cu^(+)+e^(-) rarr Cu`, will beA. `0.90 V`B. `0.30 V`C. `0.38 V`D. `0.52 V` |
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Answer» Correct Answer - D `Cu^(+2)+2e^(-) to Cu " " DeltaG_(1)^(@)` `(Cu^(+2) +e^(-) to Cu^(+) " "DeltaG_(2)^(@))/(Cu^(+) +e^(-) to Cu " " DeltaG_(3)^(@))` `DeltaG_(3)^(@) =DeltaG_(1)^(@) -DeltaG_(2)^(@)` `-1xxF xx E =- 2F xx 0.337 -(-1xxF 0.153)` `rArr E= 0.521 V` |
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