##### find the E-cell calue at 298K

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Apr 15th, 2015

Here, Ag+(aq) + e → Ag(s);

So, 2Ag+(aq) + 2e → 2Ag(s);  Eo = 0.8 V…………….. (1)

And  Cu(s)  → Cu 2+(aq) + 2e ;  Eo = -0.3402 V …………….. (2)

By, (1)+(2),

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu 2+(aq) ; Eo = (-0.3402 +0.8) = +0.4598 V

DE = 0.4598 - (0.0592/2)log ([Cu2+] / [Ag+])

= 0.4598 – 0.0296 log ([0.000850] / [0.00425])

= 0.4598 – 0.0296 *(-0.7)

= 0.4598 + 0.021

Apr 15th, 2015

the answer was incorrect. Did you take the coefficients into consideration?

Apr 15th, 2015

Sory, forgot to multiply E of Ag with 2,

Here, Ag+(aq) + e → Ag(s);

So, 2Ag+(aq) + 2e → 2Ag(s);  Eo = 1.6 V…………….. (1)

And  Cu(s)  → Cu 2+(aq) + 2e ;  Eo = -0.3402 V …………….. (2)

By, (1)+(2),

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu 2+(aq) ; Eo = (-0.3402 +1.6) = +1.2598 V

DE = 1.2598 - (0.0592/2)log ([Cu2+] / [Ag+])

= 1.2598 – 0.0296 log ([0.000850] / [0.00425])

= 1.2598 – 0.0296 *(-0.7)

= 1.2598 + 0.021

Apr 15th, 2015

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Apr 15th, 2015
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Apr 15th, 2015
Oct 24th, 2017
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