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The emf of the cell, Pt|H(2)(1atm)|H^(+)...

The emf of the cell, `Pt|H_(2)(1atm)|H^(+) (0.1M, 30mL)||Ag^(+) (0.8M)Ag` is `0.9V`. Calculate the emf when `40mL` of `0.05M NaOH` is added.

Text Solution

Verified by Experts

The correct Answer is:
`0.95V`
`E = E^(@) - 0.0591 log.([H^(+)])/([Ag^(+)])`
`0.9 - E^(@) - 0.0591 log.(0.1)/(0.8)`
`E^(@) = 0.84662`
On adding `40mL` of `0.05M NaOH`
`(0.05M NaOH 40 mL)` ltbr. `[H^(+)] = (3-2)/(70) = (1)/(70)`
now
`E = E^(@) - 0.0591 log.([H^(+)])/([Ag^(+)])`
`E = 0.84662 - 0.0591 log.(1)/(70 xx 0.8)`
`E = 0.95V`
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