Given the standard electrode potentials
`K^+//K= -2.93V, Ag^+//Ag = +0.8V, Hg^(2+)//Hg= 0.79 V, Mg^(2+)//Mg = -2.36V Cr^(3+)//Cr = 0.74 V`
Arrange them in increasing order of reducing power.

Some standard electrode potentials at 298K are given below. Cu^(2+)//Cu=0.34 V , Ca^(2+)//Ca=-2.87 V Ag^+//Ag=0.80 V , Na^+//Na= -2.71 .Write the order of metals in the decreasing order of their reducing power with proper justification.

An aqueous solution containing 1M each of Au^(3+), Cu^(2+), Ag^(+), Li^(+) is being electrolysed by using inert electrodes. The value of standard potentials are E_(Ag^(+)//Ag)^(@)= 0.80V, E_(Cu^(2+)//Cu)^(@) = 0.34V, and E_(Au^(3+)//Au)^(@) = 1.50 V , E_(Li^(+)//Li)^(@)= -3.03V , With increasing voltage, the sequence of deposition of metals on the cathode will be

N^(3-),O^(2-),F^-,Na^+, Mg^(2+),Al^(3+) Arrange them in the order of increasing ionic radii.

A list of transition metal ions are given. Ti^(2+), Sc^(3+), Cr^(2+), V^(2+), Mn^(2+), Fe^(2+), Zn^(2+):- Arrange the ions In the increasing order of their magnetic moment.

Given the standard electrod potentials E_(Cu^(2+)//Cu)^(@)=0.34v and E_(Cu^(+)//Cu)^(@)=0.52V . What is K_(c) for the disproportion 2Cu^(+)toCu+Cu^(2+)

The Standard electrode potentials of some electrodes are given below: E^@_(Zn^(2+)Zn) = -0.76 V, E^@(Cu^(2+)Cu) =+0.34, E^@_(Ag, Ag) = + 0.80V, E^@_(H, H_2)=0V What is the reaction taking place at SHE when it is connected to Ag/Ag^+. Electrode to form a galvanic cell?

Given that E values of Ag^(+) | Ag, K^(+)|K, Mg^(2+) |Mg and Cr^(3+) | Cr áre 0.80 V,-2.93 V,-2.37 V, and -0.74 V, respectively. Which of the following orders regarding the reducing power of metals is correct?