For the reaction, $\mathrm{mCuSO}_{4}+\mathrm{nKI} \rightarrow \mathrm{pCuI}+\mathrm{qI}_{2}+\mathrm

Question

For the reaction, $\mathrm{mCuSO}_{4}+\mathrm{nKI} ightarrow \mathrm{pCuI}+\mathrm{qI}_{2}+\mathrm{rK}_{2} \mathrm{SO}_{4}$\ The value of m and n in the balanced equation are respectively\

Professor PC Thomas Classes · Answer

$\stackrel{+2}{\mathrm{Cu}} \mathrm{SO}_{4}+\mathrm{K} \stackrel{-1}{\mathrm{I}} \quad ightarrow \quad \stackrel{+1}{\mathrm{Cu}} \stackrel{-1}{\mathrm{I}}+\stackrel{0}{\mathrm{I}_{2}}+\mathrm{K}_{2} \mathrm{SO}_{4}$ Since iodine is present in two products, the above redox reaction is conveniently balanced starting from the product side . Cross multiply\ $2 \mathrm{CuSO}_{4}+\mathrm{KI} ightarrow 2 \mathrm{CuI}+\mathrm{I}_{2}+2 \mathrm{~K}_{2} \mathrm{SO}_{4}$ $\mathrm{m}=2, \mathrm{n}=4$\