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$M \mid M ^{2+}$ (saturated solution of a sparingly soluble salt, $\left.MX _2\right) \| M ^{2+}\left(0.001\right.$ mol dm $\left.d ^{-3}\right) \| M$ The emf of the cell depends on the difference in concetration of $M ^{2+}$ ions at the two electrodes. The emf of the cell at $298$ is $0.059 \ V$
$1.$ The solubility product $\left( K _{ sp } ; mol ^3 dm ^{-9}\right)$ of $MX _2$ at $298$ based on the information available the given concentration cell is (take $2.303 \times R \times 298 / F =0.059 \ V$ )
$(A)$ $1 \times 10^{-15}$ $(B)$ $4 \times 10^{-15}$
$(C)$ $1 \times 10^{-12}$ $(D)$ $4 \times 10^{-12}$
$2.$ The value of $\Delta G \left( kJ \ mol ^{-1}\right)$ for the given cell is (take $1 F =96500 \ C \ mol ^{-1}$ )
$(A)$ $-5.7$ $(B)$ $5.7$ $(C)$ $11.4$ $(D)$ $-11.4$
Give the answer question $1$ and $2.$
Which forms di-iodide on reaction with $HI $ (excess) ?
