What is the potential of the cell containing two hydrogen electrodes as represented below? Pt; (1/2) H 2( g )| H 2 O || H ⊕(0.001 M )| 1 / 2 H 2( g ) Pt

Question

What is the potential of the cell containing two hydrogen electrodes as represented below? Pt; (1/2) H 2( g )| H 2 O || H ⊕(0.001 M )| 1 / 2 H 2( g ) Pt (A) -0.236 V (B) -0.0591 V (C) 0.236 V (D) 0.0591 V

Tardigrade Admin · Accepted Answer

E text cell =(0.0591/1) log ([ H ⊕] RHS /[ H ⊕] LHS ) [ text For H 2 O ,[ H ⊕]=[ overset ominusOH ]=10-7 M ] =0.059 log (10-3/10-7)=0.59 × 4=0.236 V Or E text cell =-0.059( pH c - pH )=-0.059(3-7)= 0.236 V

Q. What is the potential of the cell containing two hydrogen electrodes as represented below?
Pt; $\frac{1}{2} H_{2} (g) ∣ H_{2} O ∣ ∣ H^{\oplus} (0.001 M) ∣ 1/2 H_{2} (g) Pt$

$- 0.236 V$

$- 0.0591 V$

$0.236 V$

$0.0591 V$

$E_{cell} = \frac{0.0591}{1} lo g \frac{[ H ^{\oplus} ] _{R H S}}{[ H ^{\oplus} ] _{L H S}}$
$[For H_{2} O, [H^{\oplus}] = [O ⊖ H] = 1 0^{- 7} M]$
$= 0.059 lo g \frac{1 0 ^{- 3}}{1 0 ^{- 7}} = 0.59 \times 4 = 0.236 V$
Or $E_{cell} = - 0.059 (p H_{c} - p H) = - 0.059 (3 - 7) =$
$0.236 V$

Q. What is the potential of the cell containing two hydrogen electrodes as represented below? Pt; 21​H2​(g)∣H2​O∣∣H⊕(0.001M)∣1/2H2​(g)Pt

−0.236V

−0.0591V

0.236V

0.0591V