Two half cell reactions of an electrons of an cell are given below : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-) to Mn^(2+)(aq)+4H_(2)O(l) , E^(@)=+1.51 V` `Sn

Question

Two half cell reactions of an electrons of an cell are given below : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-) to Mn^(2+)(aq)+4H_(2)O(l) , E^(@)=+1.51 V` `Sn

Two half cell reactions of an electrons of an cell are given below :
`MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-) to Mn^(2+)(aq)+4H_(2)O(l) , E^(@)=+1.51 V`
`Sn^(2+)(aq)toSn^(4+)(aq)+2e^(-),E^(@)=+0.15 V.`
Construct a redox equation from the two half cell reactions and predict if this reaction favours the formation of reactants or products as shown in the equation.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

The cell is represented as :
`Sn^(2+)(aq)|Sn^(4+)(aq)||MnO_(4)^(-)(aq)|Mn^(2+)(aq)`
`"At cathode " : MnO_(4)^(-) (aq) +8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l)]xx2`
`{:("At anode ":,[Sn^(2+)(aq)to Sn^(4+)(aq)+2e^(-)]xx5),("overall reaction":,bar (2MnO_(4)^(-)(aq)+16H^(+)(aq)+5Sn^(2+)(aq)rarr2Mn^(2+)(aq)+5.Sn^(4+)(aq)+8H_(2)O(l))):}`
`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=^(@)MnO_(4)^(-)|Mn^(2+)-E_(Sn^(4+))^(@)|Sn^(2+)=1.51-(0.15)=1.36 V`
Since `E^(@)` cell is positive the reaction favours the formation of products.