Consider the following relations for EMF of a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The $EMF$ of a cell is defined as:$EMF_{cell} = E_{cathode} (reduction) - E_{anode} (reduction)$Since $E_{anode} (reduction) = -E_{anode} (oxidati

Vedclass · Accepted Answer

$(ii)$ and $(iv)$

Vedclass · Answer

(D) The $EMF$ of a cell is defined as:$EMF_{cell} = E_{cathode} (reduction) - E_{anode} (reduction)$Since $E_{anode} (reduction) = -E_{anode} (oxidation)$,we can write:$EMF_{cell} = E_{cathode} (reduction) + E_{anode} (oxidation)$Also,since $E_{cathode} (reduction) = -E_{cathode} (oxidation)$,we can write:$EMF_{cell} = E_{anode} (oxidation) - E_{cathode} (oxidation)$Comparing these with the given relations:$(ii)$ $EMF$ of cell = (Oxidation potential of anode) $+$ (Reduction potential of cathode) is correct.$(iv)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Oxidation potential of cathode) is correct.Therefore,the correct option is $D$.

Similar Questions

The $E^{0}_{Red}$ values of $A, B,$ and $C$ are $+0.5 \ V, -3.0 \ V,$ and $-1.2 \ V$ respectively. The order of their reducing power is:

If the $\Delta G$ of a cell reaction $AgCl + e^- \to Ag + Cl^{-}$ is $-21.20 \ kJ$,the standard e.m.f. of the cell is ............ $V$.

The standard reduction potentials for $Zn^{2+}/Zn$,$Ni^{2+}/Ni$,and $Fe^{2+}/Fe$ are $-0.76 \, V$,$-0.23 \, V$,and $-0.44 \, V$ respectively. For which of the following combinations will the reaction $X + Y^{2+} \rightarrow X^{2+} + Y$ be spontaneous?

The standard reduction electrode potentials of four elements are:
$A = -0.250 \ V$,$B = -0.136 \ V$,$C = -0.126 \ V$,$D = -0.402 \ V$
The element that displaces $A$ from its compounds is:

If $E^o_{A^{+2}/A} = -0.30 \ V$ and $E^o_{A^{+3}/A^{+2}} = 0.40 \ V$,the standard $EMF$ of the reaction: $A + 2A^{+3} \to 3A^{+2}$ will be ............ $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module