Homework Answers
Solution:- In the current scale the reference standard electrode is hydrogen electrode and the standard reduction potential for it is taken as zero and others are with respective to it.
2H+(aq) + 2e- ----------> H2(g) E0 = 0.00 V
standard reduction potential for the reduction of water in base is -0.8277 V.
2H2O(l) + 2e- ----------> H2(g) + 2OH-(aq) E0 = -0.8277 V
and the standard reduction potential for the ion M is given as +1.50 V.
M3+(aq) + 3e- -------> M(s) E0 = +1.50 V
Now, if water reduction equation is taken as the refference standard electrode and its potential is taken as 0.00 V then we could see that its potential is changing from -0.82277 V to 0.00 V means there is an incraese of 0.8277 V. So, the potential for the M must also be increased by same amount. So, new standard reduction potential for the equation, M3+(aq) + 3e- -------> M(s) will be 1.50 + 0.8277 = 2.3277 V
It could could be round to 2.33 V.
Hence, M3+(aq) + 3e- -------> M(s) E0 = +2.33 V
The answer is +2.33 V.
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electrode, with a potential of 0.242 V versus the S.H.E. Using a
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Standard potentials are measured against the standard hydrogen electrode (S.H.E.). Because it is not always...
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