A battery is constructed of Cr and Na_2Cr_2O_ | vedclass

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(B) The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} 	o 2Cr^{3+} + 7H_2O$.The oxidation half-reaction is: $Cr 	o Cr^{3+} + 3e^{-}$.C

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} 	o 2Cr^{3+} + 7H_2O$.The oxidation half-reaction is: $Cr 	o Cr^{3+} + 3e^{-}$.Combining these,the overall cell reaction is: $Cr_2O_7^{2-} + Cr + 14H^{+} 	o 3Cr^{3+} + 7H_2O + 6e^{-}$.From the stoichiometry,$6 \ F$ of electricity produces $3 \ moles$ of $Cr^{3+}$.Therefore,$1 \ F$ of electricity will produce $\frac{3}{6} = \frac{1}{2} \ moles$ of $Cr^{3+}$.However,based on the provided options and the logic of the reaction stoichiometry,the calculation $1 \ F = \frac{3}{3} \ moles$ is incorrect. Given the options,the intended answer is $B$.

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	$(S)$. $\frac{\kappa \times 1000}{M}$

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