United States presidential election in West Virginia, 1884

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United States presidential election in West Virginia, 1884

← 1880 November 4, 1884 1888 →
Turnout 21.37% of the total population Increase 3.16 pp[1]

  StephenGroverCleveland.png James G. Blaine - Brady-Handy.jpg
Nominee Grover Cleveland James G. Blaine
Party Democratic Republican
Home state New York Ohio
Running mate Thomas A. Hendricks John A. Logan
Electoral vote 6 0
Popular vote 67,311 63,096
Percentage 50.94% 47.75%

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

The 1884 United States presidential election in West Virginia took place on November 4, 1884, as part of the 1884 United States presidential election. West Virginia voters chose six representatives, or electors, to the Electoral College, who voted for president and vice president[2].

West Virginia was won by Grover Cleveland, the 28th governor of New York, (DNew York), running with the former governor of Indiana Thomas A. Hendricks, with 50.94% of the popular vote, against Secretary of State James G. Blaine (R-Ohio), running with Senator John A. Logan, with 47.75% of the vote.[2]

The Prohibition Party chose John St. John, the former governor of Kansas, and former Maryland State Senator William Daniel as their presidential and vice-presidential candidates and received 0.71% of the vote. The Greenback and Anti-Monopoly Parties both chose major general and former governor of Massachusetts Benjamin Butler and Absolom M. West, an unseated Mississippi representative, received 0.60% of the vote.

Results

United States presidential election in West Virginia, 1884[2]
Party Candidate Votes %
Democratic Grover Cleveland 67,311 50.94%
Republican James G. Blaine 63,096 47.75%
Prohibition John St. John 939 0.71%
Greenback Benjamin Butler 799 0.60%
Total votes 132,145 100.00%

References

  1. ^ "1884 Presidential Election Results West Virginia Total Population Turnout".
  2. ^ a b c "1884 Presidential Election Results West Virginia".


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