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Calculate the emf of the cell in which t...

Calculate the emf of the cell in which the reaction is
`Mg(s) + 2Ag^(+)(aq) rarr Mg^(2+) (aq) + 2Ag(s)`
when `[Mg^(2+)] = 0.130 M and [Ag^(+)] = 1.0 xx 10^(-4)M`
Given, `E_(Mg^(2+)//Mg)^(@) = -2.37 V and E_(Ag^(+)//Ag)^(@) = +0.80 V`

Text Solution

Verified by Experts

`Mg(s) rarr Mg^(2+) (aq) + 2e^(-)` (Anodic half-cell reaction)
`2Ag^(+)(aq) + 2e^(-) rarr 2Ag(s)` (Cathodic half-cell reaction)
`Mg(s) + 2Ag^(+)(aq) rarr Mg^(2+) (aq) + 2Ag(s)`
The cell may be represented as :
`Mg|Mg^(2+) (0.130 M)||Ag^(+) (1.0 xx 10^(-4)M)|Ag`
`E_(cell)^(@) = E_("Cathode")^(@) - E_("anode")^(@)`
`E_(cell)^(@) = E_(Ag^(+)//Ag)^(@) - E_(Mg^(2+)//Mg)^(@)`
= + 0.80 V - (-2.37 V)
= 0.80 V + 2.37 V = + 3.17 V
Applying Nernst equation,
`E_(cell) = E_(cell)^(@) - (0.0591)/(2)"log"([Mg^(2+)(aq)])/([Ag^(+) (aq)]^(2))" "[because [Mg(s)] = [Ag(s)] = "1 at 298 K"]`
`E_(cell) = +3.17 - (0.0591)/(2)"log" (0.130)/([1.0 xx 10^(-4)]^(2))`
`= + 3.17 - (0.0591)/(2)log (0.130 xx 10^(8))`
`= + 3.17 - 0.02955 xx (8 - (0.886)`
`= + 3.17 - 0.02955 xx 7.114`
`= + 3.17 - 0.21 = 2.96 V`
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