InterviewSolution
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InterviewSolution
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Given below are half-cell reaction: `Mn^(2+)+2e^(-) rarr Mn,, E^(@) = -1.81 V` `2(Mn^(3+)+e^(-) rarr Mn^(2+)),, E^(@) = +1.51 V` The `E^(@)` for `3Mn^(2+) rarr Mn+2Mn^(3+)` will be:A. `-0.33 V`, the reaction will not occurB. `-0.33 V`, the reaction will occurC. `-2.69 V`, the reaction will not occurD. `-2.69 V`, the reaction will occur |
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Answer» Correct Answer - C To obtain the given reaction, we reverse the second reaction and add it to the first one: `{:(Mn^(2+)+2e^(-) rarr Mn,),(2Mn^(2+)rarr2Mn^(3+)+2e^(-),):}/(3Mn^(2+)rarrMn+2Mn+2Mn^(3+))" "{:(E^(@) = -1.81 V),(E^(@) = -1.51 V):}` Thus, `E^("cell")^(@) = E_(SOP)^(@)+E_(SRP)^(@)` `= (-1.18 V)+(-1.51 V)` `= -2.69 V` Since `DeltaG_("cell")^(@) = -nFE_("cell")^(@)` negative `E_("cell")^(@)` indicates positive `DeltaG_("cell")^(@)` i.e., a non-spontaneous cell reaction. |
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