InterviewSolution
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InterviewSolution
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Two half cell reactions of an electrochemical cell are given below : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-),toMn^(2+)(aq)+4H_(2)O(l),E^(@)=+1.51V` `Sn^(2+)(aq)toSn^(4+)(aq)+2e^(-),E^(@)=+0.51V` Construct the redox equation from the two half cell reactions and predict if the reaction favours formation of reactant or product shown in the equation. |
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Answer» At anode (oxidation) : `Sn^(2+)(aq)toSn^(4+)(aq)+2e^(-),E^(@)=+0.15V` At Cathod (reduction0 : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l),E^(@)=+1.51V` Thus over all redox equation is `5Sn^(2+)(aq)+2MnO_(4)^(-)(aq)+16H^(+)(aq)to5Sn^(4+)(aq)+2Mn^(2+)(aq)+8H_(2)O(l)` `E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=0.15=1.36V` |
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