answer this plz q 11
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E° will be positive when Ag will get reduced
E°cell= E°cathode - E°anode
=0.799-0.337
=0.462
Ecell = E°cell - 0.059/n log{cathode/anode}
Ecell = 0
E°cell = 0.059/n log(Cu+2)/(Ag+)^2
0.462 = 0.059/2 log 0.1/(Ag+)^2
15.66 = log 0.1/ (Ag+)^2
solve it by taking antilog
15.66 = log 0.1 - 2 log (Ag+)^2
16.66 = -2 log (Ag+)
-8.33 = log (Ag+)
(Ag+) = 4.67 × 10^-9
I think it should be like this
E°cell= E°cathode - E°anode
=0.799-0.337
=0.462
Ecell = E°cell - 0.059/n log{cathode/anode}
Ecell = 0
E°cell = 0.059/n log(Cu+2)/(Ag+)^2
0.462 = 0.059/2 log 0.1/(Ag+)^2
15.66 = log 0.1/ (Ag+)^2
solve it by taking antilog
15.66 = log 0.1 - 2 log (Ag+)^2
16.66 = -2 log (Ag+)
-8.33 = log (Ag+)
(Ag+) = 4.67 × 10^-9
I think it should be like this
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