The voltage of a cell whose half cell reactions are given below is Mg2-2e- Mgs- E-2-37 Vanode Cu2-2e- Cus- E-0-34 Vcathod

The correct option is C 2.03 V We have E_Mg^2 + /Mg = #160; 2.37V and E_Cu^2 + /Cu = 0.34 Now #160; E_ cell ^∘ #160; = E_ C u^ 2+ / Cu ...

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Standard XII

Chemistry

Solubility Product

The voltage o...

Question

The voltage of a cell whose half cell reactions are given below is
$M g^{2 +} + 2 e^{-} \to M g (s)$ ; $E = - 2.37 V$ (anode)
$C u^{2 +} + 2 e^{-} \to C u (s)$ : $E = + 0.34 V$ (cathod)

- 2.03 V

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1.36 V

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2.71 V

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2.03 V

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Solution

The correct option is C $- 2.03 V$
We have
$E_{\frac{M g^{2 +}}{M g}} = - 2.37 V a n d E_{\frac{C u^{2 +}}{C u}} = 0.34$
Now
$\begin{matrix} E_{c e l l}^{\circ} = E_{\frac{C u^{2 +}}{C u}} - E_{\frac{M g^{2 +}}{M g}} = 0.34 - (- 2.34) = 2.71 V \end{matrix}$

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Similar questions

The voltage of a cell whose halfcells are given below is:

$M g^{2 +} + 2 e^{-} \to M g (s); E^{0} = - 2.37 V$

$C u^{2 +} + 2 e^{-} \to C u (s); E^{0} = + 0.34 V$ standard EMF of the cell is:

A l^{3 +} (a q) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{\circ} = + 0.34 V

What voltage is produced under standard conditions to give a spontaneous reactions by combination of these two-half cells?

Q. If for the half cell reactions

C u^{2 +} + e^{-} \to C u^{+}

E^{o} = 0.15 V

C u^{2 +} + 2 e^{-} \to C u

E^{o} = 0.34 V

Calculate

E^{o}

of the half cell reaction

C u^{+} + e^{-} \to C u

also predict whether

C u^{+}

undergoes disproportionation or not.

For a cell, the cell reaction is $M g (s) + C u^{2 +} (a q) \to M g^{2 +} (a q) + C u (s)$ . If the S.R.P. values of $M g$ and $C u$ are $- 2.37 V$ and $+ 0.34 V$ respectively, the EMF of the cell is:

Q. Which of the following represents the anode half-cell reaction for the following galvanic cell:

C u (s) | C u^{2 +} (a q) | | A g^{+} (a q) | A g (s)

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The voltage of a cell whose half cell reactions are given below isMg2++2e−→Mg(s); E=−2.37V(anode)Cu2++2e−→Cu(s): E=+0.34V(cathod)

The correct option is C −2.03VWe haveEMg2+Mg=−2.37VandECu2+Cu=0.34Now E∘cell=ECu2+Cu−EMg2+Mg=0.34−(−2.34)=2.71V

The voltage of a cell whose half cell reactions are given below is
$M g^{2 +} + 2 e^{-} \to M g (s)$ ; $E = - 2.37 V$ (anode)
$C u^{2 +} + 2 e^{-} \to C u (s)$ : $E = + 0.34 V$ (cathod)

The correct option is C $- 2.03 V$
We have
$E_{\frac{M g^{2 +}}{M g}} = - 2.37 V a n d E_{\frac{C u^{2 +}}{C u}} = 0.34$
Now
$\begin{matrix} E_{c e l l}^{\circ} = E_{\frac{C u^{2 +}}{C u}} - E_{\frac{M g^{2 +}}{M g}} = 0.34 - (- 2.34) = 2.71 V \end{matrix}$