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Consider the following four electrodes P = Cu2+(.0001 M/Cu(s) Q = Cu2+ (0.1 M)/Cu(s) R = Cu2+(0.01 M)/Cu(s) S = Cu2+(0.001 M)Cu(s) If the standard reduction potential of Cu2+ /Cu is +.34V the reduction potential in volts of the above electrode follows the order. |
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Answer» If the standard reduction potential of Cu2+/Cu is +.34V the reduction potential in volts of the above electrode follows the order. i.e., Q > R > P |
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