Answered By Leonardo F

Let's first work on the half-reactions. In the second reaction, there is the reduction of V and the oxidation of In. Hence, we can write their half-reactions:

 $2V^{3+}_{\left(aq\right)}+2e^-\rightarrow2V^{2+}_{\left(aq\right)}$2V³⁺_(aq)+2e⁻→2V²⁺_(aq) 

 $In^{1+}_{\left(aq\right)}\rightarrow In^{3+}_{\left(aq\right)}+2e^-$In¹⁺_(aq)→In³⁺_(aq)+2e⁻ 

In the third reaction, we have the reduction of Rh and the oxidation of Tl. Writing the half-reactions, we have:

 $2Rh^{4+}_{\left(aq\right)}+2e^-\rightarrow2Rh^{3+}_{\left(aq\right)}$2Rh⁴⁺_(aq)+2e⁻→2Rh³⁺_(aq) 

 $Tl^{^{1+}}_{\left(aq\right)}\rightarrow Tl^{3+}_{\left(aq\right)}+2e^-$Tl¹⁺_(aq)→Tl³⁺_(aq)+2e⁻ 

With this done, we can now take a look at the reduction potential of each chemical species present in the system. Since the first reaction is NOT spontaneous and it involves the oxidation of Tl and the reduction of V, we know that the reduction potential of Tl is higher than V. Since the second reaction involves the reduction of V and the oxidation of In, the reduction potential of V is higher than In. Finally, for the last reaction, we have the oxidation of Tl and the reduction of Rh. This means that the reduction potential of Rh is higher than Tl. So, in increasing order of reduction potential, we will have:

In<V<Tl<Rh

With this information, we know that the Rh can reduce (because it is the substance with the highest reduction potential) and the In can oxidize and we can write the following equation:

 $2Rh^{4+}_{\left(aq\right)}+In^{1+}_{\left(aq\right)}\rightarrow2Rh^{3+}_{\left(aq\right)}+In^{3+}_{\left(aq\right)}$2Rh⁴⁺_(aq)+In¹⁺_(aq)→2Rh³⁺_(aq)+In³⁺_(aq) 

This reaction will be spontaneous, with the Rh gaining 2 moles of electrons and the In losing 2 moles of electrons.