Tabulated thermodynamic data can be used to predict the standard potential of a cell even of it can't be measured directly. The Gibb's energy of the reaction.
$${K_2}Cr{O_4}\left( {aq} \right) + 2Ag\left( s \right) + 2FeC{l_3}\left( {aq} \right) \to A{g_2}Cr{O_4}\left( s \right) + 2FeC{l_2}\left( {aq} \right) + 2KCl\left( {aq} \right)$$ is $$ - 62.5\,kJ.mo{l^{ - 1}}$$ at 298 K .
The standard potential of $$A{g_2}Cr{O_4}/Ag,CrO_4^{2 - }$$ couple is (given $${E^0}\left( {F{e^{3 + }}/F{e^{2 + }}} \right) = + 0.77V$$)

Question

Tabulated thermodynamic data can be used to predict the standard potential of a cell even of it can't be measured directly. The Gibb's energy of the reaction.
$${K_2}Cr{O_4}\left( {aq} \right) + 2Ag\left( s \right) + 2FeC{l_3}\left( {aq} \right) \to A{g_2}Cr{O_4}\left( s \right) + 2FeC{l_2}\left( {aq} \right) + 2KCl\left( {aq} \right)$$ is $$ - 62.5\,kJ.mo{l^{ - 1}}$$ at 298 K .
The standard potential of $$A{g_2}Cr{O_4}/Ag,CrO_4^{2 - }$$ couple is (given $${E^0}\left( {F{e^{3 + }}/F{e^{2 + }}} \right) = + 0.77V$$)

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