Chemistry Electrochemistry

E°_cell = E° (cathode)-E° (anode) = 0.16-0.52=-0.36V.
lnK = nFE°/RT = 1*0.36/0.025 = 14.4. Ans is x10? ¹, so 144.

Sol. E? cell = E? (Sn²? |Sn) - E? (Cu²? |Cu)
= -0.16 - 0.34 = -0.50V
ΔG? = -nFE? cell
= -2 * 96500 * (-0.5) = 96500 J
= 96.5 kJ = 96500 J

KNO? , HCl and NaCl are strong electrolytes for these electrolytes of Λ_m with √c will be liner which can be given as
Λ_m = Λ? _m - A√c for strong electrolyte
Since given variation is not linear it has to be a weak electrolyte
CH? COOH is a weak electrolyte

$\mathrm{}1.52$

$\mathrm{}E=E_0-\frac{0.0591}{4}\mathrm{l}\mathrm{o}\mathrm{g}?{\left. [{\mathrm{H}}^{+}\right]}^4$

$\mathrm{E}=1.23+0.0591*\mathrm{p}\mathrm{H}$

$E=1.23+0.0591*\left(5\right)$

$E=1.52$

Each ion makes a definite contribution irrespective of the other ion

$\begin{array}{rr}& {\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{N}\mathrm{a}\mathrm{B}\mathrm{r}}-{\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{N}\mathrm{a}\mathrm{l}}={\mathrm{\Lambda }}_{{\mathrm{m}\mathrm{B}\mathrm{r}}^{-}}^0-{\mathrm{\Lambda }}_{{\mathrm{m}}_{{\mathrm{r}}^{-}}}^0\\ & {\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{K}\mathrm{B}\mathrm{r}}-{\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{N}\mathrm{a}\mathrm{B}\mathrm{r}}={\mathrm{\Lambda }}_{{\mathrm{m}\mathrm{K}}^{+}}^0-{\mathrm{\Lambda }}_{{\mathrm{m}\mathrm{N}\mathrm{a}}^{+}}^0\end{array}$

Λ? = K * 1000 / M = (2 * 10? * 10³) / 10? ³ = 20 Scm²mole? ¹
For weak acid (α) = Λ? / Λ? = 20 / 190 = 2/19
K? = Cα² / (1 - α) = 10? ³ * (2/19)² / (1 - 2/19)
= 10? ³ * (4/361) / (17/19) = 0.01238 * 10? ³
= 12.38 * 10?
So, ans is 12 (the nearest integer).