1892 United States presidential election in Vermont

1892 United States presidential election in Vermont

← 1888 November 8, 1892 1896 →
 
Nominee Benjamin Harrison Grover Cleveland
Party Republican Democratic
Home state Indiana New York
Running mate Whitelaw Reid Adlai Stevenson I
Electoral vote 4 0
Popular vote 37,992 16,325
Percentage 68.09% 29.26%

County Results
Harrison
  60-70%
  70-80%
  80-90%


President before election

Benjamin Harrison
Republican

Elected President

Grover Cleveland
Democratic

The 1892 United States presidential election in Vermont took place on November 8, 1892, as part of the 1892 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Vermont voted for the Republican nominee, incumbent President Benjamin Harrison, over the Democratic nominee, former President Grover Cleveland, who was running for a second, non-consecutive term. Harrison won Vermont by a margin of 38.83%.

With 68.09% of the popular vote, Vermont would be Harrison's strongest victory in terms of percentage in the popular vote.[1]

  1. ^ "1892 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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