3ni2+(aq) + 2 cr(oh)3(s) + 10 oh− (aq) → 3 ni(s) + 2 cro42−(aq) + 8 h2o(l) δg∘ = +87 kj/mol given the standard reduction potential of the half-reaction ni2+(aq) + 2 e− → ni(s) e∘red = -0.28 v, calculate the standard reduction potential of the half-reaction cro42−(aq) + 4 h2o(l) + 3 e− → cr(oh)3(s) + 5 oh−(aq)

Answer : The standard reduction potential, is -0.13 V.

Solution : Given,

      (1 KJ = 1000 J)

The net reaction is,

The half cell reactions are :

At cathode :      

At anode :  

First we have to calculate the by using formula,

where,

= Gibbs's free energy

n = number of electrons in a net chemical reaction = 6 electrons

F = Faraday constant = 96485 C

= standard cell potential

Now put all the given values in this formula, we get

Now we have to calculate the by using formula,

Now put all the given values in this formula, we get

Therefore, the standard reduction potential, is -0.13 V.

there are two significant figures.