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प्रश्न
Write the cell reaction and calculate the e.m.f of the following cell at 298 K:
`Sn(s) | Sn^(2+) (0.004 M) || H^(+) (0.020 M) | H_2 (g) ("1 bar") | Pt(s)`
(Given: `E_(Sn^(2+)"/"Sn)^0 = -0.14` V)
उत्तर
`E_"cell"^0 = E_(Sn"/"Sn^(+2))- 0.0`
= -(-0.14)
= 0.14 V
`E_"cell = E_(cell)^0 - 0.059/2 log ([H_2][Sn^(+2)])/[H^(+)]^2`
`= 0.14 - 0.059/2 log ((1)(0.004))/(0.02)^2`
`= 0.14 - 0.059/2 log (10)`
= 0.14 - 0.0295
= +0.1105 V
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