Chemistry, asked by salonigamilcom3127, 9 months ago

Calculate the emf of the following cell at 298 K 2Cr + 3Fe2+ (0.1)M ----> 2Cr3+ (0.01M) + 3Fe

E^- Cr3+ / Cr = -0.74v

E^- Fe2+/Fe = -o.44v

Answers

Answered by rashich1219

Step by step explanation:

The given cell reaction is as follows.

$2Cr(s)+3Fe^{2+}(aq)\rightarrow 2Cr^{3+}(aq)+3Fe(s)$

From the given,

$E_{Cr^{3+}/Cr}=-0.74V$

$E_{Fe^{2+}/Fe}=-0.45V$

The $E_{cell}^o$ is calculated is as follows.

$E_{cell}^o=E_{cathode}-E_{anode}$

$E_{cell}^o=-0.44V-(-0.74V)=0.3V$

$E_{cell}=E_{cell}^{o}-\frac{2303RT}{nF}log\frac{[Cr^{3+}]^{2}}{[Fe^{2+}]^{3}}$ ...................(1)

Temperature T= 298 K

n =6 (electrons involved in the reaction)

Substitute the all given values in the equation (1)

$E_{cell}=0.3V-\frac{0.0591}{6}log\frac{(0.1)^{2}}{(0.01)^{2}}V$

$E_{cell}=0.3V-\frac{0.0591}{6}log\frac{1\times10^{-2}}{1\times10^{-6}}V$

$E_{cell}=0.3V-\frac{0.0591}{6}\times 4V=0.261 V$

Therefore, Emf of the given cell is 0.261 V.