1892 United States presidential election in Rhode Island

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1892 United States presidential election in Rhode Island

← 1888 November 8, 1892 1896 →
  Pach Brothers - Benjamin Harrison.jpg StephenGroverCleveland.png
Nominee Benjamin Harrison Grover Cleveland
Party Republican Democratic
Home state Indiana New York
Running mate Whitelaw Reid Adlai Stevenson
Electoral vote 4 0
Popular vote 26,975 24,336
Percentage 50.71% 45.75%

President before election

Benjamin Harrison

Elected President

Grover Cleveland

The 1892 United States presidential election in Rhode Island 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.

Rhode Island 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 the state by a narrow margin of 4.96%.


1892 United States presidential election in Rhode Island[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Benjamin Harrison of Indiana Whitelaw Reid of New York 26,975 50.71% 4 100.00%
Democratic Grover Cleveland of New York Adlai Ewing Stevenson I of Illinois 24,336 45.75% 0 0.00%
Prohibition John Bidwell of California James Cranfill of Texas 1,654 3.11% 0 0.00%
People's James Baird Weaver of Iowa James Gaven Field of Virginia 228 0.43% 0 0.00%
N/A Others Others 3 0.01% 0 0.00%
Total 53,196 100.00% 4 100.00%


  1. ^ "1892 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.