UPSEE Solved Model Paper 1

pH = 3, [H+] = 10−3M
𝐴𝑙𝑠𝑜 𝑎𝑡 298K,

2.303RT

F

= 0.0591V
𝐸𝑙𝑒𝑡𝑟𝑜𝑛𝑠 𝑖𝑛𝑣𝑜𝑙𝑣𝑒𝑑 𝑛 = 4
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡, 𝑄 =

[Fe2+]2

[H+]−4PO2

=

(1×10−2)2

(1×10−3)4(0.1)

= 107
By Nernst equation
∴ Ecell = Ecell−

0.0591

n

log Q
= 1.67 -

0.0591

4

‌log‌107
= 1.67 − 0.10 = 1.57V