Question

Calculate emf of the cell. A|a+3(0.1 m)||b+2(0.01 m)| b given: eoa/a+3=0.75v , eob/b+2=0.45 v

Answer 1

Answer:

.26

Explanation:

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Answer 2

Answer:

The emf of the given cell will be equal to 0.26V.

Explanation:

The given cell reaction:

At anode:        $a$   →   $a^{3+}+3e^{-}$                              .........(1)

At cathode:    $b^{2+}+2e^{-}$  →   $b$                               ..........(2)

Multiple the eq. (1) by 2 and eq. (2) by 3 then add we get

        $2a+3b^{2+}$   →  $2a^{3+}+3b$

To find e.m.f. use the Nernst equation:

        $E_{cell}=E_{0}-\frac{0.059}{n}log\frac{[a^{3+}]^{2}}{[b^{2+}]^{3}}$

Standard emf :   $E^{o}=E^{o}_{c}-E^{o}_{A}$

                      $E^{o}=(-0.45)-(-0.75)=0.30V$  

Now,      $E_{cell}=E_{0}-\frac{0.059}{6}log\frac{[0.1]^{2}}{[0.01]^{3}}$

 ∴             $E_{cell}=0.30-(0.01)log10^{4}\\ E_{cell}=0.30-(10^{-2})(4)\\ E_{cell}=0.30-0.04\\ E_{cell}=0.26V$

Therefore the emf of the cell would be 0.26V.