Answer:
pH = 8.314
Explanation:
equil: S S 3S
∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24
⇒ 6.0 E-24 = ( S )*( 3S )³
⇒ 6.0 E-24 = 27S∧4
⇒ 2.22 E-25 = S∧4
⇒ ( 2.22 E-25 )∧(1/4) = S
⇒ S = 6.866 E-7 M
⇒ [ OH- ] = 3*S =2.06 E-6 M
⇒ pOH = - Log [ OH- ]
⇒ pOH = - Log ( 2.06 E-6 )
⇒ pOH = 5.686
∴ pH = 14 - pOH
⇒ pH = 8.314
Answer:
Al is oxidized while Ag is reduced.
Explanation:
The complete molecular equation is;
3Ag2S + 2Al --> 6Ag + Al2S3
Oxidation half equation;
2Al ------> 2Al^3+ + 6e
Reduction half equation;
6Ag^+ + 6e -------> 6Ag
Overall redox reaction equation;
2Al + 6Ag^+ ----->2Al^3+ + 6Ag
Hence; Al is oxidized while Ag is reduced.
