Question

What is ${E^0}$ cell if:
$2Cr + 3{H_2}O + 3OC{l^ - } \to 2C{r^{3 + }} + 3C{l^ - } + 6O{H^ - }$
$C{r^{ + 3}} + 3{e^ - } \to Cr,{E^0} = - 0.74V$
$OC{l^ - } + {H_2}O + 2{e^ - } \to C{l^ - } + 2O{H^ - },{E^0} = 0.94V$

Answer 1

Hint : Given are the cell potentials at anode and cathode. Generally, the reduction half-cell which is a cathode has a higher value than the oxidation half-cell which is an anode. The standard cell potential $\left( {{E^0}} \right)$ can be calculated from the cell potential of cathode minus the cell potential of anode.

Complete Step By Step Answer:
Cell potential gives the value about the voltage flows between the two half cells. The two half cells include the reduction half-cell and oxidation half-cell. The reduction half cell means the half cell at which the gain of electrons i.e.., reduction takes place and the oxidation half cell means the half cell at which the loss of electrons i.e.., oxidation takes place.
In the above last two reactions, both the reactions undergo gain of electrons. Thus, both undergo reduction. But which reaction has a higher ${E^0}$ value can be called a cathode and a lower value half cell can be called an anode.
$OC{l^ - } + {H_2}O + 2{e^ - } \to C{l^ - } + 2O{H^ - }$ , is the reaction which takes place at cathode.
$C{r^{ + 3}} + 3{e^ - } \to Cr$ is the reaction which takes place at anode.
Thus, ${E^0}$ cell is equal to ${E^0}_{cathode} - {E^0}_{anode}$
Substitute the values of ${E^0}_{cathode}$ which is $0.94V$ and ${E^0}_{anode}$ which is $- 0.74V$ .
The ${E^0}$ cell is $0.94V - \left( { - 0.74V} \right)$
Upon simplification, we will get ${E^0} = 1.68V$
${E^0}$ cell is $1.68V$ .

Note :
The cathode has a higher cell potential value then the cell potential of anode. Generally, cathode undergoes gain of electrons and anode undergoes loss of electrons. But in this problem both cells undergo loss of electrons. In this case the half cell with higher cell potential can be considered as a cathode.