The temperature coefficient of the e-m-f- of cell- left-displaystylefrac-dE-dT-right-p is given by-displaystylefrac-nF-triangle S-displaystylefrac-triangle S-nF-displaystylefrac-triangle S-nFT-nFE

Question

Toppr Admin · Accepted Answer

As we know-xA0-xA0- -xA0- -xA0- -xA0-xA0- -xA0- Ecell-x2212-x394-GnF-xA0- -xA0- -xA0-x394-G-x394-SdT-xA0- -xA0- The temperature coefficient -dEdT-x394-S-dTdT-nF-x394-SnFHence- option -B- is correct-

The temperature coefficient of the e.m.f. of cell, ${(\frac{d E}{d T})}_{p}$ is given by:
(A) $\frac{n F}{△ S}$
(B) $\frac{△ S}{n F}$
(C) $\frac{△ S}{n F T}$
(D) $- n F E$

As we know,
$E_{c e l l} = - \frac{Δ G}{n F}$

$Δ G = \frac{Δ S}{d T}$

The temperature coefficient $(\frac{d E}{d T}) = \frac{Δ S / d T}{d T . n F} = \frac{Δ S}{n F}$
Hence, option (B) is correct.

The temperature coefficient of the e.m.f. of cell, (dEdT)p is given by:(A) nF△S(B) △SnF(C) △SnFT(D) −nFE

As we know, Ecell=−ΔGnF ΔG=ΔSdT The temperature coefficient (dEdT)=ΔS/dTdT.nF=ΔSnFHence, option (B) is correct.

The temperature coefficient of the e.m.f. of cell, ${(\frac{d E}{d T})}_{p}$ is given by:
(A) $\frac{n F}{△ S}$
(B) $\frac{△ S}{n F}$
(C) $\frac{△ S}{n F T}$
(D) $- n F E$

As we know,
$E_{c e l l} = - \frac{Δ G}{n F}$

$Δ G = \frac{Δ S}{d T}$

The temperature coefficient $(\frac{d E}{d T}) = \frac{Δ S / d T}{d T . n F} = \frac{Δ S}{n F}$
Hence, option (B) is correct.