Determine the unknown concentration of the ion in each of thefollowing cells. (a

Question

Determine the unknown concentration of the ion in each of thefollowing cells. (a) Pb(s) |Pb2+(aq, ?) || Pb2+(aq,0.10 mol·L-1) | Pb(s), E =+0.033 V.
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(b) Pt(s) | Fe2+(aq, 1.0mol·L-1) , Fe3+(aq, 0.10mol·L-1) || Fe3+(aq, ?),Fe2+(aq, 0.0010 mol·L-1) |Pt(s), E = +0.12 V.

Express answer in M.
2
(a) Pb(s) |Pb2+(aq, ?) || Pb2+(aq,0.10 mol·L-1) | Pb(s), E =+0.033 V.
1

(b) Pt(s) | Fe2+(aq, 1.0mol·L-1) , Fe3+(aq, 0.10mol·L-1) || Fe3+(aq, ?),Fe2+(aq, 0.0010 mol·L-1) |Pt(s), E = +0.12 V.

Express answer in M.
2

Explanation / Answer

      According to Nernestequation :          E ( cell) = E0 (cell) - 0.059 / n log [oxidation ] / [ Reduction ]             Oxidationhalf cell :
          Pb (s)------> Pb+2 (aq. ? M ) + 2 e          Reductionhalf cell :         Pb+2 ( aq. , 0.10 M ) + 2 e ------> Pb (s)        E (cell) = 0 - 0.059 /  2 log [ Pb+2 ] / [ 0.10 M ]          0.033 V = - 0.059 / 2 log [ Pb+2 ] / [ 0.10 M ]              Pb+2   = 0.00760 M                  0.033 V = - 0.059 / 2 log [ Pb+2 ] / [ 0.10 M ]              Pb+2   = 0.00760 M b.   Oxidation half - cell :          Fe+2 ( aq. 1.0 M )  ------->   Fe+3 (aq. 0.10 M ) + e       Reduction half - cell :          Fe+3 (aq. ? M) + e --------> Fe+2 ( aq. 0.0010 M )                   E(cell) = 0 - 0.059 /  1 log [0.10 M ] [ 0.0010 M ] / [ 1.0 M ] [ ? M ]          0.12 V = - 0.059 / 1 log [ 0.10 M ] [ 0.0010M ] / [ 1.0 M ] [ Fe +3 ]              Fe+3   = 0.0108 M

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