Standard electrode potential for $S{n^{4 + }}$ $|S{n^{2 + }}$ couple is $ + 0.15V $ and that for the $C{r^{3 + }}$ $|Cr$ couple is $ - 0.74V$. These two couples in their standard state are connected to make a cell potential will be :
A. $ + 1.19V$
B. $ + 0.89V$
C. $ + 0.18V$
D. $ + 1.83V$

Hint:Standard electrode potential is calculated by half reaction of each process involved in the reaction. The overall standard potential is formed by the addition of cathode and anode electrode potential involved in the reaction.

Complete answer:
standard electrode potential is defined as the measure the individual potential of reversible electrode at standard state with ions at an effective concentration of $1mold{m^{ - 3}}$ at the pressure of $1$ atm. All electrochemical cells are based on redox reactions, which are made up of two half-reactions. The oxidation half-reaction occurs at the anode and it involves a loss of electrons. Reduction reaction takes place at the cathode, involving a gain of electrons. The electric potential that arises between the anode and the cathode is due to the difference in the individual potentials of each electrode. The cell potential of an electrochemical cell can be measured with the help of a voltmeter.
${E_{Sn^{4+}/Sn^{2+}}} = +0.15V $
${E_{(Cr^{3 +} /Cr)}} = - 0.74V$
The electrode potential is calculated by difference in the electrode potential of redox reaction involved at cathode and anode.
Now,
$E{^\circ _{cell}} = {E_c}^\circ - {E_A}^\circ $
$
   {E^{\circ}}_{cell} = 0.15 - ( - 0.74)
    $
${E^{\circ}}_{cell} = + 0.89V
$
The standard electrode is $ + 0.89V$

Therefore the correct option is B.

Note:
The oxidation potential of an electrode is the negative of its reduction potential. Therefore, the standard electrode potential of an electrode is described by its standard reduction potential. Good oxidizing agents have high standard reduction potentials whereas good reducing agents have low standard reduction potentials.