JEE Main 2019 (Online) 8th April Evening Slot | Electrochemistry Question 82 | Chemistry

JEE Main 2019 (Online) 8th April Evening Slot

MCQ (Single Correct Answer)

-1

Calculate the standard cell potential in (V) of the cell in which following reaction takes place : Fe²⁺(aq) + Ag⁺(aq) $$ \to $$ Fe³⁺(aq) + Ag (s)

Given that
$$E_{A{g^ + }/Ag}^o = xV$$
$$E_{Fe^{2+ }/Fe}^o = yV$$
$$E_{Fe^{3+ }/Fe}^o = zV$$

x + 2y - 3z

x - z

x - y

x + y - z

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

Given that $${E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V$$ ;
$${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V$$
$${E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V$$
$${E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V$$

The strongest oxidizing agent is :

O₂

Au³⁺

Br₂

$${S_2}O_8^{2 - }$$

JEE Main 2019 (Online) 12th January Evening Slot

MCQ (Single Correct Answer)

-1

$$ \wedge _m^ \circ $$ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm² mol^–1, respectively. If the conductivity of 0.001 M HA is-

5 $$ \times $$ 10^–5 S cm^–1, degree of dissociation of HA is -

0.50

0.125

0.25

0.75

JEE Main 2019 (Online) 12th January Evening Slot

MCQ (Single Correct Answer)

-1

Given

(i)  C (graphite) + O₂(g) $$ \to $$ CO₂(g); $$\Delta $$rH$$^\Theta $$ = x kJ mol^$$-$$1

(ii)  C(graphite) + $${1 \over 2}$$O₂(g) $$ \to $$ CO₂(g); $$\Delta $$rH$$^\Theta $$ = y kJ mol^$$-$$1

(iii)  CO(g) + $${1 \over 2}$$ O₂(g) $$ \to $$ CO₂(g); $$\Delta $$rH$$^\Theta $$ = z kJ mol^$$-$$1

Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct ?

z = x + y

x = y + z

x = y – z

y = 2z – x

PREVIOUS NEXT