Question

A solution containing one mole per litre each of $AX,B{X}_2,C{X}_2$ and $D{X}_2$ is electrolysed using inert electrodes. The values of the standard potentials for reduction reactions of $A^{+}|A,B^{2+}|B,C^{2+}|C$ and $D^{2+}|D$ are $+0.80,+0.34,-0.76$ and $-1.66$ volts respectively.
The correct sequence in which these metals will be deposited on the cathode is:

Answer 1

As we are given

$A^{+}/A=+0.80$

$B^{2+}/B==0.34$

$C^{2+}/C=-0.76$

$D^{2+}/D=-1.66$

Now the sequence will be $A>B>C>D$

because the deposition on the cathode will be from the maximum value of ions to the lowest one.