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Calculate the standard cell potential of the galvanic cell in which the following reaction takes place : 2Cr(s) + 3Cd^(2+) (aq) to 2Cr^(3+) (aq) + 3Cd (s) Also calculate the Delta_(r)G^(@) value of the reaction. [Given: E_(Cr^(3+)//Cr)^(@) = -0.74 V: E_(Cd^(2+)//Cd)^(@) = -0.40 V and F = 96500 C mol^(-1)] |
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Answer» Solution : The electrode REACTIONS may be represented as under : `2Cr (s) to 2Cr^(3+)(aq) + 6e^(-)` `3Cd^(2+)(aq) + 6e^(-) to 3Cd(s)` Thus, n =6 Standard cell potential may be OBTAINED as under : `E_("cell")^(@) = E_(Cd^(2+)//Cd)^(@) -E_(Cr^(3+)//Cr)^(@) =-0.40 V -(-0.74 V)` or `E_("cell")^(@) =0.34 V` To calculate `Delta_(r)G^(@)` , applying the following RELATION and substituting the VALUES, we get: `Delta_(r)G^(@) =-nE^(@) F = -6 xx 0.34 V xx 96500 C` or `Delta_(r)G^(@) = -196.86 KJ mol^(-1)` |
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