Question

plz answer this guys

I will mark you as a brainlest
(chapter electrochemistry)​

Answer 1

Answer:

(i) Mg(s)/Mg²⁺ (0.001M) || Cu²⁺ (0.0001M)/Cu(s)

At anode : Mg(s) ⇢ Mg²⁺(aq) + 2e⁻

At cathode : Cu²⁺(aq) + 2e⁻ ⇢ Cu(s)

Overall reaction : Mg(s) + Cu²⁺(aq) ⇢ Mg²⁺(aq) + Cu(s)

As we know,

${\sf{\bigstar \ \ E_{cell} = E^\degree _{cell} - {\dfrac{0.059}{2}} log Q_c}}$

Here, ${\sf{Q_c = {\dfrac{[Mg^{2+}]}{[Cu^{2+}]}} }}$

Nernst equation :

${\sf{E_{cell} = E^\degree _{cell} - {\dfrac{0.059}{2}} log {\dfrac{[Mg^{2+}]}{[Cu^{2+}]}} }}$

Now,

EMF of the cell :

${\sf{E^\degree _{cell} = E^\degree _{cathode} - E^\degree _{anode} }}$

${\sf{E^\degree _{cell} = E^\degree _{Cu^{2+} (aq)/Cu(s)} - E^\degree _{Mg^{2+} (aq)/Mg(s)} }}$

= 0.33 - (- 2.36)

= 0.34 + 2.36

= 2.70 V

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(ii) Pt(s)/Br₂(l)/Br⁻(0.010M) || H⁺(0.030M)/H₂(g) 1bar/Pt(s)

At anode : 2Br⁻(aq) ⇢ Br₂(l) + 2e⁻

At cathode : 2H⁺(aq) + 2e⁻ ⇢ H₂(g)

Overall reaction : 2Br⁻(aq) + 2H⁺(aq)⇢ Br₂(l) + H₂(g)

As we know,

${\sf{\bigstar \ \ E_{cell} = E^\degree _{cell} - {\dfrac{0.059}{2}} log Q_c}}$

Here, ${\sf{Q_c = {\dfrac{1}{[Br^-]^2[H^+]^2}} }}$

Nernst equation :

${\sf{E_{cell} = E^\degree _{cell} - {\dfrac{0.059}{2}} log {\dfrac{1}{[Br^-]^2[H^+]^2}} }}$

Now,

EMF of the cell :

${\sf{E^\degree _{cell} = E^\degree _{cathode} - E^\degree _{anode} }}$

${\sf{E^\degree _{cell} = E^\degree _{H^+ / H_2} - E^\degree _{Br_2 / Br^-} }}$

= 0 - 1.09

= - 1.09 V