The easiest way to work this is to determine the Eocell, then correct for the concentrations not being standard of 1 M. You've made some typos. I assume you meant to write:
when Fe^2+ = 0.8102 M
Ni^2+ = 0.3310 M
Fe^2+ + 2e = -0.4459
N^2+ +2e E= -0.2520
Fe(s) ==> Fe^2+ + 2e ......Eo = +0.4459 v
Ni^2+ + 2e ==> Ni(s).........Eo = -0.2520 v
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Fe(s) + Ni^2+ ==> Ni(s) + Fe^2+ ......Eocell = 0.1939 v
Then plug Eocell and Qrxn into the following and solve for Ecell at the concentrations of the problem.
Ecell = Eocell - 0.05916/n log Q
where n = 2 and
Q = (Ni)(Fe^2+)/(Fe)(Ni^2+). For Q, substitute 0.8102 M for Fe^2+ and 0.3310 M for Ni^2+ as in the problem. By definition (Fe) and (Ni) = 1.000.
Post your work if you get stuck.
Voltage prroduced
calculate the voltage produced by the voltaic cell between fe and ni^2+
when Fe = 0.8102M
Ni^2+ = 0.3310M
Fe +2e E= -0.4459
Ni +2e E= -0.2520
1 answer
1 answer