Consider a voltaic cell in which the following reaction takesplace: 2Fe2+(aq)+H2

Question

Consider a voltaic cell in which the following reaction takesplace:
2Fe2+(aq)+H2O2(aq)+2H+(aq)-->2Fe3+(aq)+2H2O What is the cell potential of this cell operating with[Fe2+]=0.00813 M; [H2O2]=0.914M; [Fe3+]=0.199M; and a pHof2.88?
2Fe2+(aq)+H2O2(aq)+2H+(aq)-->2Fe3+(aq)+2H2O What is the cell potential of this cell operating with[Fe2+]=0.00813 M; [H2O2]=0.914M; [Fe3+]=0.199M; and a pHof2.88? Consider a voltaic cell in which the following reaction takesplace:
2Fe2+(aq)+H2O2(aq)+2H+(aq)-->2Fe3+(aq)+2H2O What is the cell potential of this cell operating with[Fe2+]=0.00813 M; [H2O2]=0.914M; [Fe3+]=0.199M; and a pHof2.88?

Explanation / Answer

We Know that :     Oxidation - half cell :        Fe+2   --------> Fe+3+ e   ] x 2     Reduction - half cell :       H2O2 + 2 H+ +2e ------> 2 H2O          E0cell = E0cathode - E0anode                     = + 1.78 V - ( + 0.77 ) V                      = + 1.01 V          According to Nernest equation :         Ecell = E0cell -0.059 / 2 log [ Fe+3 ]2 / [Fe+2 ] 2 [H2O2] [ H+]2     pH = 2.88 ; [ H+ ] = 0.00131 M                    = 1.01 V - 0.059 / 2 log [ 0.199 M]2 / [ 0.00813 M ]2 [ 0.941 M] [ 0.00131 M ]2                     =  0.7572 V                     =  0.7572 V

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