Why is electrochemistry important for the invention of the latest technology?

Consider the strong electrolytes $Z_{m}X_{n}$,$U_{m}Y_{p}$ and $V_{m}X_{n}$. Limiting molar conductivity $(\Lambda^0)$ of $U_{m}Y_{p}$ and $V_{m}X_{n}$ are $250 \ S \ cm^2 \ mol^{-1}$ and $440 \ S \ cm^2 \ mol^{-1}$,respectively. The value of $(m + n + p)$ is . . . . . Given:

$Ion$	$\lambda^0 \ (S \ cm^2 \ mol^{-1})$
$U^{p+}$	$50.0$
$Y^{m-}$	$50.0$
$V^{n+}$	$60.0$
$X^{m-}$	$50.0$
$Z^{n+}$	$40.0$

$\lambda^0$ is the limiting molar conductivity of ions. The plot of molar conductivity $(\Lambda)$ of $Z_{m}X_{n}$ $vs$ $c^{1/2}$ is given below.

Given below are the half-cell reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn; E^{o} = -1.18 \ V$
$2(Mn^{3+} + e^{-} \rightarrow Mn^{2+}); E^{o} = +1.51 \ V$
The $E^{o}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be:

$Cu^{+}$ ion is not stable in aqueous solution because of disproportionation reaction. $E^o$ value for disproportionation of $Cu^{+}$ is .............. $V$ (Given $E^o_{Cu^{2+}/Cu^{+}} = 0.15 \ V$,$E^o_{Cu^{2+}/Cu} = 0.34 \ V$)

The molar conductivity of $0.025 \ mol \ L^{-1}$ methanoic acid is $46.1 \ S \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.

Column $I$	Column $II$
$(A)$. Kohlrausch law can calculate	$(P)$. $\frac{\Lambda_m^c}{\Lambda_m^o}$
$(B)$. Molar conductance $\Lambda_m$	$(Q)$. $\frac{1}{R} \times \frac{l}{A}$
$(C)$. Specific conductance $\kappa$	$(R)$. $\Lambda_m^o$ of $Ca_3(PO_4)_2$
$(D)$. Degree of ionization of weak electrolyte	$(S)$. $\frac{\kappa \times 1000}{M}$

Which of the following options shows the correct matches?