For the given voltaic cell Mn(s) + Cd2+(aq) <==> Mn2+(aq) + Cd(s) (a) initial Ec

Question

For the given voltaic cell

Mn(s) + Cd2+(aq) <==> Mn2+(aq) + Cd(s)

(a) initial Ecell

Nernst equation,

Ecell = Eo - 0.0592/n log([Mn2+]/[Cd2+])

         = (-0.40 - (- 1.18)) - 0.0592/2 log(0.090/0.060)

         = 0.775 V

(b) after [Cd2+] reached = 0.035 M

[Mn2+] = 0.090 + (0.060 - 0.035) = 0.115 M

Ecell = Eo - 0.0592/n log([Mn2+]/[Cd2+])

         = (-0.40 - (- 1.18)) - 0.0592/2 log(0.115/0.035)

         = 0.765 V

(c) when Ecell = 0.055 V

Ecell = Eo - 0.0592/n log([Mn2+]/[Cd2+])

0.055 = (-0.40 - (- 1.18)) - 0.0592/2 log[(0.090 + x)/(0.060 -x)]

1.87 x 10^23 - 3.113 x 10^24x = 0.090 + x

x = 1.87 x 10^23/3.113 x 10^24 = 0.060 M

So,

[Mn2+] = 0.090 + 0.060 = 0.150 M

(d) Equilibrium concentration of ions at Ecell = 0.055 V

[Mn2+] = 0.150 M

[Cd2+] = 0 M

Explanation / Answer

3 atempts left Check my work Be sure to answer all parts. A voltaic cell with Mn/Mn2+ and Cd/Cd2+ half-cells has the following initial concentrations: [Mn2+)-0.090 M; [Cd2+] = 0.060 M. (a) What is the initial E ,? cell 0.77 V (b) What is E when [Cd2+] reaches 0.035 M? (e Whatis (Mn+I when E eli reaches 0.055 v? (d) What are the equilibrium concentrations of the ions? (Enter your answer in scientifie notation.) [Cd2+]=-1x10 M [Mn2+ ] = M

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