Giúp em với ạ

Đáp án: 

a,

(1)$CrO_4^{2-}+2H^++2H_2O+3e \to [Cr(OH)_6]^{3-} \varepsilon^o_{CrO_4^{2-}/[Cr(OH)_6]^{3-}}=-0,16V$ 

$\varepsilon_{CrO_4^{2-}/[Cr(OH)_6]^{3-}}=\varepsilon^o_{CrO_4^{2-}/[Cr(OH)_6]^{3-}}-\dfrac{RT}{nF}\ln \dfrac{1}{[H^+]^2}=-0,16-\dfrac{8,314.298}{3.96485}\ln \dfrac{1}{10^{-18}}=-0,195V$ 

(2) $Na_2O_2+2H_2O+2e \to 2Na^++4OH^-$ 

$\varepsilon^o_{Na_2O_2/OH^-, Na^+}=1,2V$  

$\varepsilon_{Na_2O_2/OH^-, Na^+}=\varepsilon^o_{Na_2O_2/OH^-, Na^+}-\dfrac{RT}{nF}\ln [OH^-]^4=1,2-\dfrac{8,314.298}{2.96485}\ln 10^{-5.4}=1,791V$ 

b,

$\varepsilon^o_{Na_2O_2/OH^-, Na^+}>\varepsilon^o_{CrO_4^{2-}/[Cr(OH)_6]^{3-}}$ nên phản ứng xảy ra là:

(3) $3Na_2O_2+2[Cr(OH)_3]^{3-}\to 6Na^++2OH^- + 2CrO_4^{2-}+ 2H_2O$

c,

$\varepsilon^o_{(3)}=\varepsilon^o_{(2)}-\varepsilon^o_{(1)}=1,2-(-0,16)=1,36V$ 

$\Delta G^o_{(3)}=-nF\varepsilon^o_{(3)}=-6.96485.1,36=-787317,6 J=-787,3kJ$  

$\Delta G^o_{(3)}=-RT\ln K\Rightarrow K=\text{exp}\left[\dfrac{787317,6}{8,314.298}\right]=e^{317,78}$